11 Projectile Motion Practice Problems With Answers

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. 1. Marble off a bench

   A marble rolls off a level lab bench 1.20 m above the floor, leaving the edge horizontally at $2.5 \ \mathrm{m/s}$. How long is it in the air, and how far from the base of the bench does it land?

   Show the worked solution

   1. Take up as positive and the launch point as the origin. Because the marble leaves horizontally, $v_{y0} = 0$ and $v_{x0} = 2.5 \ \mathrm{m/s}$. The floor is at $y = -1.20 \ \mathrm{m}$.

   2. Time, from the vertical column. With $v_{y0} = 0$, $y = v_{y0} t - \frac{1}{2} g t^2$ becomes $-1.20 = -\frac{1}{2}(9.8) t^2 = -4.9 t^2$.

   3. Solve: $t^2 = \frac{1.20}{4.9} = 0.24490 \ \mathrm{s^2}$, so $t = 0.4949 \ \mathrm{s}$, or $0.495 \ \mathrm{s}$ to three figures.

   4. Range, from the horizontal column. $x = v_{x0} t = (2.5 \ \mathrm{m/s})(0.4949 \ \mathrm{s}) = 1.237 \ \mathrm{m}$, or $1.24 \ \mathrm{m}$.

   5. Note what did not appear in the time calculation: the launch speed. A marble leaving at $2.5 \ \mathrm{m/s}$ and one leaving at $25 \ \mathrm{m/s}$ hit the floor at the same instant. The faster one just lands ten times farther out.

   The marble is in the air for $0.495 \ \mathrm{s}$ and lands $1.24 \ \mathrm{m}$ from the base of the bench.

2. 2. Horizontal launch off a cliff, with impact velocity

   A rock is kicked horizontally at $8.5 \ \mathrm{m/s}$ off the edge of a cliff 62.0 m above flat ground. Find the time of flight, the distance from the base of the cliff where it lands, and its velocity at impact as a magnitude and a direction.

   Show the worked solution

   1. Take up as positive with the launch point as the origin. Horizontal: $v_{x0} = 8.5 \ \mathrm{m/s}$, $a_x = 0$. Vertical: $v_{y0} = 0$, $a_y = -9.8 \ \mathrm{m/s^2}$, landing at $y = -62.0 \ \mathrm{m}$.

   2. Time. $-62.0 = -4.9 t^2$, so $t^2 = \frac{62.0}{4.9} = 12.653 \ \mathrm{s^2}$ and $t = 3.5571 \ \mathrm{s}$, which is $3.56 \ \mathrm{s}$ to three figures. Keep the extra digits for the next two parts.

   3. Range. $x = v_{x0} t = (8.5)(3.5571) = 30.24 \ \mathrm{m}$, or $30.2 \ \mathrm{m}$.

   4. Impact components. Horizontal velocity never changed, so $v_x = 8.5 \ \mathrm{m/s}$. Vertical: $v_y = 0 - (9.8)(3.5571) = -34.86 \ \mathrm{m/s}$. Independent check with the no-time equation: $v_y^2 = 0 + 2(9.8)(62.0) = 1215.2$, so $|v_y| = 34.86 \ \mathrm{m/s}$. Agreed.

   5. Magnitude. $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(8.5)^2 + (34.86)^2} = \sqrt{72.25 + 1215.2} = \sqrt{1287.5} = 35.88 \ \mathrm{m/s}$, or $35.9 \ \mathrm{m/s}$.

   6. Direction. $\tan\phi = \frac{|v_y|}{v_x} = \frac{34.86}{8.5} = 4.101$, so $\phi = 76.3^\circ$ below the horizontal. The rock is falling far more steeply than it is travelling forward, which is what a 62 m drop against a gentle kick should produce.

   Flight time $3.56 \ \mathrm{s}$, landing $30.2 \ \mathrm{m}$ from the base, impact velocity $35.9 \ \mathrm{m/s}$ directed $76.3^\circ$ below the horizontal.

3. 3. Angled launch from level ground

   A golf ball is struck from level ground at $32.0 \ \mathrm{m/s}$, $28^\circ$ above the horizontal, and lands on ground at the same height. Find the time of flight, the range, and the maximum height.

   Show the worked solution

   1. Take up as positive, launch point as the origin. Components: $v_{x0} = 32.0 \cos 28^\circ = 32.0(0.8829) = 28.25 \ \mathrm{m/s}$ and $v_{y0} = 32.0 \sin 28^\circ = 32.0(0.4695) = 15.02 \ \mathrm{m/s}$.

   2. Time of flight. Launch and landing heights match, so the flight is symmetric: the ball takes as long coming down as it took going up, and $t = \frac{2 v_{y0}}{g} = \frac{2(15.02)}{9.8} = 3.065 \ \mathrm{s}$, or $3.07 \ \mathrm{s}$.

   3. Range. $R = v_{x0} t = (28.25)(3.065) = 86.59 \ \mathrm{m}$, or $86.6 \ \mathrm{m}$.

   4. Maximum height. At the top $v_y = 0$, so $0 = v_{y0}^2 - 2 g H$ gives $H = \frac{v_{y0}^2}{2g} = \frac{(15.02)^2}{19.6} = \frac{225.6}{19.6} = 11.51 \ \mathrm{m}$, or $11.5 \ \mathrm{m}$.

   5. Cross-check the range with the level-ground shortcut: $R = \frac{v_0^2 \sin 2\theta}{g} = \frac{(32.0)^2 \sin 56^\circ}{9.8} = \frac{1024(0.8290)}{9.8} = 86.6 \ \mathrm{m}$. It agrees here because launch and landing heights are equal, which is the only case in which it is safe.

   Time of flight $3.07 \ \mathrm{s}$, range $86.6 \ \mathrm{m}$, maximum height $11.5 \ \mathrm{m}$.

4. 4. Speed at the top of the arc

   A ball is launched from ground level at $19.0 \ \mathrm{m/s}$, $70^\circ$ above the horizontal. Find the time to reach the highest point, the maximum height, and the ball's speed at that highest point. Explain why the speed there is not zero.

   Show the worked solution

   1. Take up as positive. Components: $v_{x0} = 19.0 \cos 70^\circ = 19.0(0.3420) = 6.498 \ \mathrm{m/s}$ and $v_{y0} = 19.0 \sin 70^\circ = 19.0(0.9397) = 17.85 \ \mathrm{m/s}$.

   2. Time to the peak. The peak is defined by $v_y = 0$. From $v_y = v_{y0} - g t$: $0 = 17.85 - 9.8 t$, so $t = \frac{17.85}{9.8} = 1.821 \ \mathrm{s}$, or $1.82 \ \mathrm{s}$.

   3. Maximum height. $H = \frac{v_{y0}^2}{2g} = \frac{(17.85)^2}{19.6} = \frac{318.6}{19.6} = 16.26 \ \mathrm{m}$, or $16.3 \ \mathrm{m}$.

   4. Speed at the peak. Only the vertical component is zero. The horizontal component was never touched by gravity, so it is still $6.498 \ \mathrm{m/s}$, and the speed at the top is $\sqrt{(6.498)^2 + 0^2} = 6.50 \ \mathrm{m/s}$, directed horizontally.

   5. Why it is not zero: gravity points straight down, so it has no horizontal component and can never change $v_x$. Calling the whole velocity zero at the peak would mean the ball stops in mid-air and then falls straight down, which is not the shape of the trajectory. The one thing that is zero at the peak is $v_y$, and the one thing that is not zero anywhere is the acceleration, still $9.8 \ \mathrm{m/s^2}$ downward at that instant.

   Peak at $t = 1.82 \ \mathrm{s}$, maximum height $16.3 \ \mathrm{m}$, and a speed there of $6.50 \ \mathrm{m/s}$ directed horizontally. Only the vertical component of the velocity vanishes at the top.

5. 5. The velocity vector partway through the flight

   A stone is launched from level ground at $26.0 \ \mathrm{m/s}$, $55^\circ$ above the horizontal. Find its velocity $3.20 \ \mathrm{s}$ after launch, as a magnitude and a direction, and say whether the stone is rising or falling at that moment.

   Show the worked solution

   1. Take up as positive. Components at launch: $v_{x0} = 26.0 \cos 55^\circ = 26.0(0.5736) = 14.91 \ \mathrm{m/s}$ and $v_{y0} = 26.0 \sin 55^\circ = 26.0(0.8192) = 21.30 \ \mathrm{m/s}$.

   2. First check the stone is still in the air. The flight is symmetric about the peak at $t = \frac{21.30}{9.8} = 2.173 \ \mathrm{s}$, so the total flight lasts $4.347 \ \mathrm{s}$. At $3.20 \ \mathrm{s}$ the stone is past the peak and still airborne.

   3. Horizontal component. Unchanged, because $a_x = 0$: $v_x = 14.91 \ \mathrm{m/s}$.

   4. Vertical component. $v_y = v_{y0} - g t = 21.30 - (9.8)(3.20) = 21.30 - 31.36 = -10.06 \ \mathrm{m/s}$. The minus sign says downward, consistent with being past the peak.

   5. Magnitude. $v = \sqrt{(14.91)^2 + (10.06)^2} = \sqrt{222.3 + 101.2} = \sqrt{323.5} = 17.99 \ \mathrm{m/s}$, or $18.0 \ \mathrm{m/s}$.

   6. Direction. $\tan\phi = \frac{10.06}{14.91} = 0.6747$, so $\phi = 34.0^\circ$ below the horizontal. Quote both numbers: a magnitude with no direction is not a velocity.

   7. Sense check against the launch: the speed has dropped from $26.0$ to $18.0 \ \mathrm{m/s}$ and the direction has swung from $55^\circ$ above the horizontal to $34.0^\circ$ below it. Both changes came entirely from the vertical component.

   $18.0 \ \mathrm{m/s}$ directed $34.0^\circ$ below the horizontal. The stone is falling, having passed its peak at $2.17 \ \mathrm{s}$.

6. 6. Clearing a wall

   A ball is launched from ground level at $21.0 \ \mathrm{m/s}$, $60^\circ$ above the horizontal, straight toward a wall 6.0 m tall whose base is 30.0 m from the launch point. Does the ball clear the wall, and by how much?

   Show the worked solution

   1. Take up as positive, launch point as the origin. Components: $v_{x0} = 21.0 \cos 60^\circ = 21.0(0.5000) = 10.50 \ \mathrm{m/s}$ and $v_{y0} = 21.0 \sin 60^\circ = 21.0(0.8660) = 18.19 \ \mathrm{m/s}$.

   2. The question is about a position, not a landing, so do not compute the range first. Find the instant the ball is level with the wall, then find how high it is at that instant.

   3. Time to reach the wall. From the horizontal column: $t = \frac{x}{v_{x0}} = \frac{30.0}{10.50} = 2.857 \ \mathrm{s}$.

   4. Height at that instant. $y = v_{y0} t - \frac{1}{2} g t^2 = (18.19)(2.857) - (4.9)(2.857)^2 = 51.97 \ \mathrm{m} - 40.00 \ \mathrm{m} = 11.97 \ \mathrm{m}$, which is $12.0 \ \mathrm{m}$ to three figures.

   5. Verdict. $12.0 \ \mathrm{m}$ against a wall of $6.0 \ \mathrm{m}$, so the ball passes about $6.0 \ \mathrm{m}$ above the top.

   6. Worth confirming the ball actually gets that far. Level-ground range: $R = \frac{(21.0)^2 \sin 120^\circ}{9.8} = \frac{441(0.8660)}{9.8} = 39.0 \ \mathrm{m}$, comfortably past the wall at 30.0 m. Had the range come out under 30 m, the ball would have landed short and the height calculation would have been meaningless.

   Yes. The ball is $12.0 \ \mathrm{m}$ up when it reaches the wall, clearing the $6.0 \ \mathrm{m}$ top by about $6.0 \ \mathrm{m}$.

7. 7. Angled launch from a roof: the unsymmetric case

   A ball is thrown from the edge of a roof 18.0 m above flat ground, at $14.0 \ \mathrm{m/s}$, $35^\circ$ above the horizontal. Find the time until it hits the ground, the horizontal distance from the base of the building where it lands, its maximum height above the ground, and its impact speed.

   Show the worked solution

   1. Take up as positive with the launch point as the origin, so the ground is at $y = -18.0 \ \mathrm{m}$. Components: $v_{x0} = 14.0 \cos 35^\circ = 14.0(0.8192) = 11.47 \ \mathrm{m/s}$ and $v_{y0} = 14.0 \sin 35^\circ = 14.0(0.5736) = 8.030 \ \mathrm{m/s}$.

   2. Symmetry is gone here, because the ball lands lower than it started. The trip down is longer than the trip up, so $t = 2 v_{y0}/g$ does not apply and you must solve the full quadratic.

   3. Time. $-18.0 = 8.030 t - 4.9 t^2$, which rearranges to $4.9 t^2 - 8.030 t - 18.0 = 0$.

   4. Quadratic formula: $t = \frac{8.030 \pm \sqrt{(8.030)^2 + 4(4.9)(18.0)}}{2(4.9)} = \frac{8.030 \pm \sqrt{64.48 + 352.8}}{9.8} = \frac{8.030 \pm 20.43}{9.8}$. The minus root is negative, meaning a time before the throw, so keep $t = \frac{28.46}{9.8} = 2.904 \ \mathrm{s}$, or $2.90 \ \mathrm{s}$.

   5. Range. $x = v_{x0} t = (11.47)(2.904) = 33.31 \ \mathrm{m}$, or $33.3 \ \mathrm{m}$ from the base.

   6. Maximum height above the ground. The rise above the launch point is $\frac{v_{y0}^2}{2g} = \frac{(8.030)^2}{19.6} = \frac{64.48}{19.6} = 3.290 \ \mathrm{m}$, so the peak sits $18.0 + 3.29 = 21.3 \ \mathrm{m}$ above the ground. The question asked from the ground, so the roof height has to be added back on.

   7. Impact speed. $v_y = 8.030 - (9.8)(2.904) = -20.43 \ \mathrm{m/s}$, and $v_x$ is still $11.47 \ \mathrm{m/s}$, so $v = \sqrt{(11.47)^2 + (20.43)^2} = \sqrt{131.6 + 417.3} = \sqrt{548.9} = 23.4 \ \mathrm{m/s}$.

   8. Independent check on the impact speed, using the no-time equation on the vertical column and then recombining: $v^2 = v_0^2 + 2 g h = (14.0)^2 + 2(9.8)(18.0) = 196 + 352.8 = 548.8$, so $v = 23.43 \ \mathrm{m/s}$. It matches, and notice the launch angle dropped out: any $14.0 \ \mathrm{m/s}$ throw from that roof lands at the same speed, though not at the same place or angle.

   Flight time $2.90 \ \mathrm{s}$, landing $33.3 \ \mathrm{m}$ from the base, peak $21.3 \ \mathrm{m}$ above the ground, impact speed $23.4 \ \mathrm{m/s}$.

8. 8. Complementary angles give the same range

   Two balls are launched from level ground at the same speed, $24.0 \ \mathrm{m/s}$: the first at $25^\circ$ above the horizontal, the second at $65^\circ$. Find the range, flight time, and peak height of each, and explain the pattern in the numbers.

   Show the worked solution

   1. Take up as positive for both. Note that $25^\circ$ and $65^\circ$ add to $90^\circ$, so $\cos 65^\circ = \sin 25^\circ$ and $\sin 65^\circ = \cos 25^\circ$. The two launches have the same pair of components, swapped between the axes.

   2. Ball A at $25^\circ$. $v_{x0} = 24.0 \cos 25^\circ = 24.0(0.9063) = 21.75 \ \mathrm{m/s}$ and $v_{y0} = 24.0 \sin 25^\circ = 24.0(0.4226) = 10.14 \ \mathrm{m/s}$. Symmetric flight, so $t = \frac{2(10.14)}{9.8} = 2.069 \ \mathrm{s}$, or $2.07 \ \mathrm{s}$. Range $R = (21.75)(2.069) = 45.01 \ \mathrm{m}$, or $45.0 \ \mathrm{m}$. Peak $H = \frac{(10.14)^2}{19.6} = 5.25 \ \mathrm{m}$.

   3. Ball B at $65^\circ$. $v_{x0} = 24.0 \cos 65^\circ = 24.0(0.4226) = 10.14 \ \mathrm{m/s}$ and $v_{y0} = 24.0 \sin 65^\circ = 24.0(0.9063) = 21.75 \ \mathrm{m/s}$. Symmetric flight, so $t = \frac{2(21.75)}{9.8} = 4.439 \ \mathrm{s}$, or $4.44 \ \mathrm{s}$. Range $R = (10.14)(4.439) = 45.01 \ \mathrm{m}$, or $45.0 \ \mathrm{m}$. Peak $H = \frac{(21.75)^2}{19.6} = 24.1 \ \mathrm{m}$.

   4. The pattern. Ball B is in the air $2.15$ times as long and rises $4.6$ times as high, yet the two ranges are identical, because B's horizontal speed is smaller by exactly the factor that its flight time is larger. Written out: $R = v_{x0} t = (v_0\cos\theta)\left(\frac{2 v_0 \sin\theta}{g}\right) = \frac{v_0^2 \sin 2\theta}{g}$, and $\sin(2 \times 25^\circ) = \sin 50^\circ$ equals $\sin(2 \times 65^\circ) = \sin 130^\circ$.

   5. So on level ground, any pair of angles adding to $90^\circ$ gives the same range at the same launch speed. The flat trajectory gets there sooner; the steep one hangs. The single angle with no partner is $45^\circ$, which is why it is the maximum: $\sin 2\theta$ peaks at $2\theta = 90^\circ$.

   6. Both results depend on the flights being symmetric, so both depend on the launch and landing heights matching. Problem 11 shows what breaks when they do not.

   Both ranges are $45.0 \ \mathrm{m}$. The $25^\circ$ ball takes $2.07 \ \mathrm{s}$ and peaks at $5.25 \ \mathrm{m}$; the $65^\circ$ ball takes $4.44 \ \mathrm{s}$ and peaks at $24.1 \ \mathrm{m}$. Complementary angles on level ground always tie on range.

9. 9. Solving for an unknown launch angle

   A projectile is launched from level ground at $30.0 \ \mathrm{m/s}$ and lands 78.0 m away, at the same height. Find both launch angles that produce this range, and say which one puts the projectile in the air longer.

   Show the worked solution

   1. Take up as positive. Launch and landing heights match, so the flight is symmetric and the level-ground range relation applies: $R = \frac{v_0^2 \sin 2\theta}{g}$. This relation is not printed on the AP sheet, so derive it in one line if you need it, as in problem 8.

   2. Rearrange for the angle: $\sin 2\theta = \frac{R g}{v_0^2}$.

   3. Substitute: $\sin 2\theta = \frac{(78.0)(9.8)}{(30.0)^2} = \frac{764.4}{900} = 0.8493$.

   4. A sine equation has two solutions in the relevant range, and this is where marks are usually lost. $2\theta = 58.14^\circ$ or $2\theta = 180^\circ - 58.14^\circ = 121.86^\circ$.

   5. So $\theta = 29.07^\circ$ or $\theta = 60.93^\circ$, that is $29.1^\circ$ and $60.9^\circ$. They add to $90.0^\circ$, the complementary pair from problem 8.

   6. Check the shallow one the long way: $v_{y0} = 30.0 \sin 29.07^\circ = 14.58 \ \mathrm{m/s}$, $v_{x0} = 30.0 \cos 29.07^\circ = 26.22 \ \mathrm{m/s}$, flight time $t = \frac{2(14.58)}{9.8} = 2.975 \ \mathrm{s}$, range $(26.22)(2.975) = 78.0 \ \mathrm{m}$. Back to the given range.

   7. Flight times: the steep launch has $v_{y0} = 30.0 \sin 60.93^\circ = 26.22 \ \mathrm{m/s}$ and $t = \frac{2(26.22)}{9.8} = 5.35 \ \mathrm{s}$, against $2.98 \ \mathrm{s}$ for the shallow one. Same landing spot, very different hang time.

   8. One more thing to notice: $\sin 2\theta$ cannot exceed 1, so this launch speed cannot reach beyond $\frac{(30.0)^2}{9.8} = 91.8 \ \mathrm{m}$ on level ground. If a problem asks for a range past that limit, the arcsine has no solution and the honest answer is that the shot is impossible at that speed.

   $\theta = 29.1^\circ$ or $\theta = 60.9^\circ$. The steep launch stays airborne $5.35 \ \mathrm{s}$ against $2.98 \ \mathrm{s}$ for the shallow one, and both land 78.0 m away.

10. 10. Why the two directions can be treated separately

    Explain why horizontal and vertical motion can be solved independently for a projectile, and what the two solutions have in common. Then use the idea to answer this: a ball is dropped from a window at the same instant an identical ball is thrown horizontally from that window. Which lands first?

    Show the worked solution

    1. Start from the force. Once the object is in flight, the only force on it is gravity, which points straight down. A vector pointing straight down has zero horizontal component.

    2. Feed that into Newton's second law, one axis at a time. Horizontally there is no force, so $a_x = 0$ and $v_x$ is constant for the whole flight. Vertically the force is $mg$ downward, so $a_y = -g$, independent of mass. That is the CED's definition of projectile motion in essential knowledge 1.5.B.2: zero acceleration in one dimension and constant, nonzero acceleration in the other.

    3. Because neither equation contains a variable belonging to the other axis, they are two separate one-dimensional problems. Essential knowledge 1.5.B.1 makes exactly this point: motion in two dimensions can be analyzed with one-dimensional kinematic relationships once the motion is separated into components.

    4. The one thing the two columns share is the clock. Both describe the same object, so the same value of $t$ appears in both. That shared variable is what lets you solve the vertical column for the time and hand it to the horizontal one, and it is the only bridge between them.

    5. Now the two balls. They leave the same window at the same instant, so both have $y_0$ equal and $v_{y0} = 0$: the throw was horizontal, so it put nothing into the vertical column. Their vertical equations are therefore character for character identical, $y = y_0 - \frac{1}{2} g t^2$, and produce the same landing time.

    6. The thrown ball travels sideways the whole way down, so it lands farther out, but not later. If the window is 5.0 m up, both land at $t = \sqrt{\frac{2(5.0)}{9.8}} = 1.01 \ \mathrm{s}$, whatever the horizontal throw speed was.

    7. The usual objection is that the thrown ball is moving faster, so surely it takes longer. It is moving faster, but the extra speed is entirely horizontal, and horizontal speed appears nowhere in the vertical equation.

    Neither. Gravity has no horizontal component, so $a_x = 0$ and $a_y = -g$ are separate equations linked only by the shared time. Both balls start with $v_{y0} = 0$ from the same height, so their vertical motion is identical and they land together; the thrown one simply lands farther from the wall.

11. 11. When 45 degrees is not the best angle

    The standard result is that $45^\circ$ maximizes the range of a projectile. Explain the condition that result depends on. Then test it: a ball is launched at $20.0 \ \mathrm{m/s}$ from a platform 15.0 m above flat ground. Compare the range at $45^\circ$ with the range at $35^\circ$.

    Show the worked solution

    1. The condition is that the projectile lands at the same height it launched from. Under that condition the flight is symmetric, $R = \frac{v_0^2 \sin 2\theta}{g}$ holds, and the range is largest when $\sin 2\theta = 1$, that is $\theta = 45^\circ$. Lift the launch point above the landing point and the symmetry breaks, the formula stops applying, and the best angle drops below $45^\circ$.

    2. The reason is a trade-off. Raising the angle buys hang time but costs horizontal speed. Launching from a height gives you free extra hang time on the way down that you did not have to pay for with angle, so the optimum shifts toward keeping more of the launch speed horizontal.

    3. Take up as positive with the launch point as the origin, so the ground is at $y = -15.0 \ \mathrm{m}$ and every flight time comes from $4.9 t^2 - v_{y0} t - 15.0 = 0$.

    4. At $45^\circ$. $v_{x0} = v_{y0} = 20.0 \cos 45^\circ = 14.14 \ \mathrm{m/s}$. Then $t = \frac{14.14 + \sqrt{(14.14)^2 + 4(4.9)(15.0)}}{9.8} = \frac{14.14 + \sqrt{199.9 + 294}}{9.8} = \frac{14.14 + 22.22}{9.8} = 3.711 \ \mathrm{s}$, and $R = (14.14)(3.711) = 52.5 \ \mathrm{m}$.

    5. At $35^\circ$. $v_{x0} = 20.0 \cos 35^\circ = 16.38 \ \mathrm{m/s}$ and $v_{y0} = 20.0 \sin 35^\circ = 11.47 \ \mathrm{m/s}$. Then $t = \frac{11.47 + \sqrt{(11.47)^2 + 294}}{9.8} = \frac{11.47 + \sqrt{131.6 + 294}}{9.8} = \frac{11.47 + 20.63}{9.8} = 3.275 \ \mathrm{s}$, and $R = (16.38)(3.275) = 53.7 \ \mathrm{m}$.

    6. The shallower launch wins by about $1.2 \ \mathrm{m}$, despite spending $0.44 \ \mathrm{s}$ less in the air, because it carries $2.24 \ \mathrm{m/s}$ more horizontal speed for every second of that flight. Scanning angles between the two would find the best around $37^\circ$ for these numbers, worth roughly $53.8 \ \mathrm{m}$.

    7. The same argument run the other way says that launching into a higher landing point pushes the best angle above $45^\circ$. The safe habit is to treat $45^\circ$ as the answer to one specific question, level ground and no air resistance, rather than as a fact about projectiles.

    $45^\circ$ maximizes range only when the launch and landing heights are equal. From a 15.0 m platform at $20.0 \ \mathrm{m/s}$, the $35^\circ$ launch reaches $53.7 \ \mathrm{m}$ against $52.5 \ \mathrm{m}$ at $45^\circ$, and the true optimum is near $37^\circ$.