The problems are on the projectile motion worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Projectile motion is two-dimensional motion under gravity alone. These worked answers correct the 20 problems of the AP Physics 1 worksheet on projectile motion, from components of a bat strike to planning a launcher test: each solution shows the formula, the substitution and the final answer with its unit.
Also, each answer shows the formula, the substitution and the final result in bold with its unit. Compare your method as well as your numbers, because AP scoring rewards the reasoning.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 1
Problem 1 Components of a bat strike
1. vx = 30 cos 20° = 28.2 m/s.2. So, vy = 30 sin 20° = 10.3 m/s.
3. Thus, The horizontal component stays constant because ax = 0.
Problem 2 True or false at the top
1. True: vy = 0 at the top.2. Therefore, False: vx is unchanged.
3. Hence, False.
4. Then, True: a = 9.81 m/s² downward throughout the flight.
Problem 3 Walk across a field
1. distance = 40 + 30 = 70 m.2. Displacement = √(40² + 30²) = √2500 = 50 m.
3. Thus, tan θ = 30 ÷ 40 = 0.75, so θ = 36.9° north of east.
Problem 4 Marble leaving a table
1. 0.80 = ½ × 9.81 × t², so t = √(1.6 ÷ 9.81) = 0.40 s.2. x = 1.5 × 0.404 = 0.61 m.
Problem 5 Dropped or launched
1. They land together.2. Hence, Both start with zero vertical velocity and have the same vertical acceleration and the same height, so y = ½gt² gives the same time. In addition, the horizontal motion does not affect the vertical motion.
Problem 6 Football over level ground
1. vx = 25 cos 40° = 19.2 m/s; vy0 = 25 sin 40° = 16.1 m/s.2. T = 2 × 16.07 ÷ 9.81 = 3.28 s.
3. R = 19.15 × 3.276 = 62.7 m.
4. Thus, H = 16.07² ÷ (2 × 9.81) = 13.2 m.
Problem 7 Reading a velocity graph
1. Also, the graph crosses zero at 1.5 s.2. Gradient = (−14.7 − 14.7) ÷ 3.0 = −9.8 m/s²; it is the acceleration, equal to −g (within rounding).
3. Hence, Area of the triangle = ½ × 1.5 × 14.7 = 11 m.
4. Then, the ball lands at 3.0 s, when vy = −14.7 m/s.
Problem 8 Stone thrown from a cliff
1. 45 = ½ × 9.81 × t², t = √(90 ÷ 9.81) = 3.03 s.2. x = 8.0 × 3.03 = 24 m.
3. Thus, vy = 9.81 × 3.03 = 29.7 m/s; speed = √(8.0² + 29.7²) = 30.8 m/s.
4. tan θ = 29.7 ÷ 8.0, so θ = 75° below the horizontal.
Problem 9 Moving walkway
1. 1.4 + 0.80 = 2.2 m/s.2. Hence, t = 66 ÷ 2.2 = 30 s.
3. Then, 1.4 − 0.80 = 0.60 m/s in his walking direction.
Problem 10 Spot the error
1. vx = 20 cos 60° = 10 m/s.2. So, At the top vy = 0, so the speed equals vx = 10 m/s.
3. Thus, Only the vertical component is zero at the top. The horizontal component never changes, so the speed is not zero.
Problem 11 Free kick and crossbar
1. vx = 18 cos 35° = 14.75 m/s; t = 20 ÷ 14.75 = 1.36 s.2. Therefore, vy0 = 18 sin 35° = 10.32 m/s; y = 10.32 × 1.356 − ½ × 9.81 × 1.356² = 14.0 − 9.0 = 5.0 m.
3. Hence, 5.0 m is above 2.44 m, so the ball passes over the bar.
Problem 12 Same speed, different angles
1. R = 15² × sin 90° ÷ 9.81 = 22.9 m.2. R = 225 × sin 60° ÷ 9.81 = 19.9 m.
3. R = 225 × sin 120° ÷ 9.81 = 19.9 m. Angles that add up to 90° give the same range, because sin 2θ is equal for both.
Problem 13 River crossing
1. t = 80 ÷ 2.0 = 40 s.2. Thus, Drift = 1.5 × 40 = 60 m.
3. v = √(2.0² + 1.5²) = 2.5 m/s.
Problem 14 Reading a strobe diagram
1. Gravity acts only vertically, so ax = 0 and the ball covers equal distances in equal times: 10 × 0.5 = 5 m per interval.2. Hence, y = ½ × 9.81 × 1.0² = 4.9 m, which matches the diagram (4.91 m).
3. y = ½ × 9.81 × 2.5² = 31 m.
Problem 15 Ball hitting a wall
1. vx = 20 cos 50° = 12.86 m/s; t = 18 ÷ 12.86 = 1.40 s.2. So, vy0 = 20 sin 50° = 15.32 m/s; y = 15.32 × 1.40 − 4.905 × 1.40² = 21.4 − 9.6 = 11.8 m.
3. Thus, vy = 15.32 − 9.81 × 1.40 = +1.6 m/s, positive, so the ball is still rising.
Problem 16 Stone thrown from a bridge
1. Afterwards, take up as positive and the water as y = −12 m: −12 = 7.5t − 4.905t² (vy0 = 15 sin 30° = 7.5 m/s). So 4.905t² − 7.5t − 12 = 0, giving t = (7.5 + 17.08) ÷ 9.81 = 2.51 s.2. Therefore, vx = 15 cos 30° = 12.99 m/s; x = 12.99 × 2.506 = 32.5 m.
3. Hence, vy = 7.5 − 9.81 × 2.506 = −17.1 m/s; speed = √(12.99² + 17.08²) = 21.5 m/s at tan⁻¹(17.08 ÷ 12.99) = 53° below the horizontal.
Problem 17 Heading into the wind
1. Ground velocity components: 15 m/s east and 60 m/s north, so speed = √(15² + 60²) = 62 m/s, at tan⁻¹(15 ÷ 60) = 14° east of north.2. Then, the plane must cancel the 15 m/s eastward drift: sin θ = 15 ÷ 60 = 0.25, so θ = 14.5° west of north.
3. Ground speed = √(60² − 15²) = 58 m/s due north.
Problem 18 Basketball from its peak
1. vy0 = √(2gH) = √(2 × 9.81 × 3.2) = 7.9 m/s.2. Thus, v = 7.92 ÷ sin 60° = 9.1 m/s.
3. T = 2 × 7.92 ÷ 9.81 = 1.62 s; vx = 9.15 cos 60° = 4.57 m/s; R = 4.57 × 1.62 = 7.4 m.
Problem 19 Analysing tracking data
1. vx = 2.4 ÷ 0.2 = 12 m/s (constant for every row).2. Hence, y ÷ t = 8.0, 7.05, 6.05, 5.08 m/s. Since y ÷ t = vy0 − ½gt, a line through these points extrapolates to t = 0 at about 9.0 m/s. Moreover, the gradient (−4.9 m/s²) confirms ½g.
3. v = √(12² + 9.0²) = 15 m/s; θ = tan⁻¹(9.0 ÷ 12) = 37°.
4. t = 9.0 ÷ 9.81 = 0.92 s.
Problem 20 Planning a launcher test
1. Also, clamp the launcher at the bench edge, fire the ball several times, and use a plumb line to mark the point on the floor below the edge. In addition, lay white paper with carbon paper on top where the ball lands, and measure the horizontal range x and the bench height h with the rule.2. Thus, t = √(2h ÷ g) and v = x ÷ t, so v = x √(g ÷ 2h).
3. Afterwards, repeat at least ten times and average the range; tape the paper down; measure from the plumb line to the centre of the cluster of marks.
4. Therefore, Fall time would increase by a factor √2, so the range would be √2 times larger (1.41×), while the launch speed stays the same.
