Vertical Motion Practice Problems AP Physics 1 — 8 Problems with Solutions
Vertical motion is one of the most tested kinematics topics on AP Physics 1. These 8 practice problems cover free fall, objects launched upward, and projectile motion — with full step-by-step solutions showing exactly how to earn maximum credit on the AP exam.
Key Equations for Vertical Motion
For all vertical motion problems, the acceleration is $a = -g = -9.8$ m/s² (taking up as positive). The four kinematic equations apply:
$$v = v_0 + at$$ $$y = y_0 + v_0 t + \frac{1}{2}at^2$$ $$v^2 = v_0^2 + 2a\Delta y$$ $$\bar{v} = \frac{v + v_0}{2}$$
Sign convention: Up is positive. Initial upward velocity is positive. Acceleration due to gravity is negative (−9.8 m/s²). Displacement upward is positive, downward is negative.
Problem 1 — Free Fall from Rest (Easy)
A ball is dropped from rest from a height of 45 m. How long does it take to hit the ground? What is its speed just before impact?
Solution:
Take the ground as $y = 0$. Initial position $y_0 = 45$ m, initial velocity $v_0 = 0$, $a = -9.8$ m/s².
For time: $y = y_0 + v_0 t + \frac{1}{2}at^2$
$$0 = 45 + 0 - \frac{1}{2}(9.8)t^2$$ $$t^2 = \frac{45}{4.9} = 9.18$$ $$\boxed{t = 3.03 \text{ s}}$$
For final speed: $v^2 = v_0^2 + 2a\Delta y = 0 + 2(-9.8)(-45) = 882$
$$\boxed{v = 29.7 \text{ m/s (downward)}}$$
Problem 2 — Object Thrown Upward (Easy)
A ball is thrown straight upward with an initial speed of 20 m/s.
(a) How high does it go? (b) How long until it returns to the starting height?
Solution:
At maximum height, $v = 0$. $v_0 = +20$ m/s, $a = -9.8$ m/s².
(a) $v^2 = v_0^2 + 2a\Delta y$
$$0 = (20)^2 + 2(-9.8)\Delta y$$ $$\Delta y = \frac{400}{19.6} = \boxed{20.4 \text{ m}}$$
(b) By symmetry, the ball takes the same time going up as coming down. Time to reach max height:
$$0 = 20 + (-9.8)t \Rightarrow t = 2.04 \text{ s}$$
Total time = $2 \times 2.04 = \boxed{4.08 \text{ s}}$
Problem 3 — Object Thrown Upward from a Height (Medium)
A ball is thrown upward at 15 m/s from the top of a 30 m cliff. How long does it take to hit the ground below? What is its speed at impact?
Solution:
Take the ground as $y = 0$. $y_0 = 30$ m, $v_0 = +15$ m/s, $a = -9.8$ m/s².
$$0 = 30 + 15t - \frac{1}{2}(9.8)t^2$$ $$4.9t^2 - 15t - 30 = 0$$
Using the quadratic formula: $t = \frac{15 \pm \sqrt{225 + 588}}{9.8} = \frac{15 \pm \sqrt{813}}{9.8}$
$\sqrt{813} \approx 28.5$
Taking the positive root: $t = \frac{15 + 28.5}{9.8} = \boxed{4.44 \text{ s}}$
Speed at impact: $v = v_0 + at = 15 + (-9.8)(4.44) = 15 - 43.5 = -28.5$ m/s
$$\boxed{|,v,| = 28.5 \text{ m/s downward}}$$
Problem 4 — Maximum Height and Time in Air (Medium)
A rocket is launched straight up and accelerates at $15$ m/s² for 5 seconds, then the engine shuts off. How high does the rocket go in total? How long is it in the air?
Solution:
Phase 1 — engine on (0 to 5 s): $v_0 = 0$, $a = +15$ m/s²
Height gained: $y_1 = \frac{1}{2}(15)(5)^2 = 187.5$ m
Speed at engine cutoff: $v_1 = 0 + (15)(5) = 75$ m/s
Phase 2 — engine off (free flight upward): $v_0 = 75$ m/s, $a = -9.8$ m/s²
Additional height: $\Delta y = \frac{v^2 - v_0^2}{2a} = \frac{0 - 75^2}{2(-9.8)} = \frac{5625}{19.6} = 287.2$ m
Total height: $187.5 + 287.2 = \boxed{474.7 \text{ m}}$
Time in Phase 2 (up): $t_2 = \frac{0 - 75}{-9.8} = 7.65$ s
Phase 3 — falling from 474.7 m: $0 = 474.7 - \frac{1}{2}(9.8)t_3^2 \Rightarrow t_3 = \sqrt{\frac{474.7}{4.9}} = 9.84$ s
Total air time: $5 + 7.65 + 9.84 = \boxed{22.5 \text{ s}}$
Problem 5 — Projectile Launched Horizontally (Medium)
A ball rolls off a table at 4.0 m/s horizontally and lands 1.2 m from the base of the table. How high is the table?
Solution:
Horizontal: $x = v_{0x} t \Rightarrow 1.2 = 4.0 \cdot t \Rightarrow t = 0.30$ s
Vertical (starting from rest vertically): $\Delta y = \frac{1}{2}g t^2 = \frac{1}{2}(9.8)(0.30)^2 = 0.44$ m
$$\boxed{h = 0.44 \text{ m}}$$
Problem 6 — Projectile at an Angle (Medium)
A ball is kicked at 25 m/s at 37° above the horizontal. Find: (a) max height, (b) horizontal range, (c) speed at max height.
Solution:
Components: $v_{0x} = 25\cos37° = 20$ m/s, $v_{0y} = 25\sin37° = 15$ m/s
(a) At max height, $v_y = 0$: $$\Delta y = \frac{v_y^2 - v_{0y}^2}{2(-9.8)} = \frac{0 - 225}{-19.6} = \boxed{11.5 \text{ m}}$$
(b) Time to max height: $t_{up} = \frac{15}{9.8} = 1.53$ s. Total time = $2 \times 1.53 = 3.06$ s
$$R = v_{0x} \cdot t_{total} = 20 \times 3.06 = \boxed{61.2 \text{ m}}$$
(c) At max height, $v_y = 0$, so only horizontal component remains: $$\boxed{v = v_{0x} = 20 \text{ m/s}}$$
Problem 7 — Two Objects, Same Drop (Hard)
Ball A is dropped from rest at height $H$. At the exact same moment, Ball B is launched straight upward from the ground with speed $v_0$. At what height do they meet?
Solution:
Let $y = 0$ at the ground, $t = 0$ at release.
Ball A: $y_A = H - \frac{1}{2}gt^2$
Ball B: $y_B = v_0 t - \frac{1}{2}gt^2$
Set equal: $H - \frac{1}{2}gt^2 = v_0 t - \frac{1}{2}gt^2$
The $\frac{1}{2}gt^2$ terms cancel: $H = v_0 t \Rightarrow t = \frac{H}{v_0}$
Height at meeting: $y = v_0 \cdot \frac{H}{v_0} - \frac{1}{2}g\left(\frac{H}{v_0}\right)^2 = H - \frac{gH^2}{2v_0^2}$
$$\boxed{y_{meet} = H\left(1 - \frac{gH}{2v_0^2}\right)}$$
Key insight: The gravity terms cancel when both objects experience the same acceleration. This is a common AP Physics 1 FRQ setup.
Problem 8 — FRQ Style: Justify and Calculate (Hard)
A student drops a ball from height $h$ and measures the time $t$ for it to reach the ground. The student claims the speed just before impact is $v = \sqrt{2gh}$.
(a) Derive this result from kinematics. (b) A second student says the result can also be derived from energy conservation. Show this. (c) If $h = 20$ m, what is the speed at impact?
Solution:
(a) Kinematic derivation: Starting from rest ($v_0 = 0$), falling distance $h$:
$$v^2 = v_0^2 + 2a\Delta y = 0 + 2(g)(h)$$ $$\boxed{v = \sqrt{2gh}}$$
(b) Energy derivation: At the top, $KE = 0$, $PE = mgh$. At the bottom, $PE = 0$, $KE = \frac{1}{2}mv^2$.
$$mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} \checkmark$$
(c) $v = \sqrt{2(9.8)(20)} = \sqrt{392} = \boxed{19.8 \text{ m/s}}$
AP Exam Tips for Vertical Motion Problems
Always define your sign convention explicitly. Write "up is positive" or "down is positive" at the start of your solution. AP readers give intent credit even if you make a sign error, as long as your convention is stated and consistent.
Use the right equation. Match the equation to what's unknown and what's given. If you have no time and want to find speed, use $v^2 = v_0^2 + 2a\Delta y$.
At maximum height, $v_y = 0$ — not $v = 0$. For projectiles, horizontal speed is unchanged. Only the vertical component is zero at peak height.
Quadratic setups are common. When an object is thrown up and lands below its launch point, you'll need the quadratic formula. Don't forget to take the positive root.