Projectile Motion — IISER Physics MCQs with Solutions
Free IISER Physics Projectile Motion MCQs with step-by-step solutions (20 questions). Part of Motion in 2D. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Projectile Motion · easy · numerical
A ball is projected from level ground at 25 m/s at 30° above the horizontal. What is its time of flight? (g = 10 m/s²)
A. 2 s
B. 5 s
C. 1.25 s
D. 2.5 s ✓ Correct
Solution: T = 2u sin θ / g = 2 × 25 × 0.5 / 10 = 2.5 s.
Q2 — Projectile Motion · easy · numerical
A sphere is launched at 20 m/s at 30° above the horizontal. What maximum height does it reach? (g = 10 m/s²)
A. 15 m
B. 5 m ✓ Correct
C. 10 m
D. 20 m
Solution: H = u² sin²θ / 2g = 400 × 0.25 / 20 = 5 m.
Q3 — Projectile Motion · easy · numerical
A robot launcher fires a projectile at 20 m/s at 45° above the horizontal. What is its range on level ground? (g = 10 m/s²)
A. 20 m
B. 40 m ✓ Correct
C. 80 m
D. 60 m
Solution: R = u² sin 2θ / g = 400 × sin 90° / 10 = 40 m.
Q4 — Projectile Motion · easy · numerical
A probe is projected at 30 m/s at 60° above the horizontal. What is its speed at the highest point of its path?
A. 15 m/s ✓ Correct
B. 30 m/s
C. 0 m/s
D. 15√3 ≈ 26 m/s
Solution: At the top the vertical component is zero, leaving only the horizontal component u cos θ = 30 × cos 60° = 15 m/s.
Q5 — Projectile Motion · easy · numerical
A sphere rolls off the edge of a 125 m high cliff, leaving it horizontally. How long does it take to reach the ground? (g = 10 m/s²)
A. 25 s
B. 2.5 s
C. 5 s ✓ Correct
D. 12.5 s
Solution: The fall is governed only by gravity: 125 = ½ × 10 × t², so t² = 25 and t = 5 s.
Q6 — Projectile Motion · easy · numerical
A ball is thrown horizontally at 15 m/s from the top of an 80 m tower. How far from the base of the tower does it land? (g = 10 m/s²)
A. 30 m
B. 80 m
C. 45 m
D. 60 m ✓ Correct
Solution: Fall time: 80 = 5t² gives t = 4 s; horizontal distance = 15 × 4 = 60 m.
Q7 — Projectile Motion · easy · numerical
A probe is launched horizontally at 16 m/s from a 45 m high platform. With what speed does it hit the ground? (g = 10 m/s²)
A. 46 m/s
B. 30 m/s
C. 16 m/s
D. 34 m/s ✓ Correct
Solution: Fall time: 45 = 5t² gives t = 3 s, so v_y = 10 × 3 = 30 m/s; impact speed = √(16² + 30²) = 34 m/s.
Q8 — Projectile Motion · easy · numerical
A trolley-mounted launcher fires a ball at 45° and it lands 62.5 m away on level ground. What was the launch speed? (g = 10 m/s²)
A. 12.5 m/s
B. 25 m/s ✓ Correct
C. 50 m/s
D. 35 m/s
Solution: At 45°, R = u²/g, so u = √(Rg) = √(62.5 × 10) = √625 = 25 m/s.
Q9 — Projectile Motion · medium · numerical
A ball is projected at 40 m/s at 60° above the horizontal. What is its time of flight? (g = 10 m/s²)
A. 8 s
B. 2√3 ≈ 3.5 s
C. 4√3 ≈ 6.9 s ✓ Correct
D. 4 s
Solution: T = 2u sin θ / g = 2 × 40 × (√3/2) / 10 = 4√3 ≈ 6.9 s.
Q10 — Projectile Motion · medium · numerical
A sphere is projected at 30 m/s at 45° above the horizontal. What maximum height does it reach? (g = 10 m/s²)
A. 11.25 m
B. 22.5 m ✓ Correct
C. 45 m
D. 30 m
Solution: H = u² sin²θ / 2g = 900 × 0.5 / 20 = 22.5 m.
Q11 — Projectile Motion · medium · numerical
A probe is fired from level ground at 50 m/s at 30° above the horizontal. What is its horizontal range? (g = 10 m/s²)
A. 125 m
B. 250 m
C. 125√3 ≈ 217 m ✓ Correct
D. 108 m
Solution: R = u² sin 2θ / g = 2500 × sin 60° / 10 = 250 × (√3/2) = 125√3 ≈ 217 m.
Q12 — Projectile Motion · medium · numerical
A ball is projected at 20 m/s at 60° above the horizontal. What is the ratio of its maximum height to its horizontal range? (g = 10 m/s²)
A. 1/2 = 0.50
B. √3/2 ≈ 0.87
C. 1/4 = 0.25
D. √3/4 ≈ 0.43 ✓ Correct
Solution: H = 400 × 0.75 / 20 = 15 m and R = 400 × (√3/2) / 10 = 20√3 ≈ 34.6 m, so H/R = tan θ / 4 = √3/4 ≈ 0.43.
Q13 — Projectile Motion · medium · numerical
A robot fires two identical spheres at 36 m/s, one at 30° and one at 60° above the horizontal, and they land at the same spot. What is the ratio of the maximum height of the first to that of the second? (g = 10 m/s²)
A. 1 : 3 ✓ Correct
B. 1 : 2
C. 1 : 9
D. 1 : √3
Solution: Complementary angles give equal ranges, while H ∝ sin²θ: H₃₀/H₆₀ = 0.25/0.75 = 1/3 (numerically 16.2 m and 48.6 m).
Q14 — Projectile Motion · medium · numerical
A drone launcher fires a probe with initial velocity 15 î + 30 ĵ (m/s), where ĵ points vertically up. What is the speed of the probe 1 s after launch? (g = 10 m/s²)
A. 35 m/s
B. 25 m/s ✓ Correct
C. 20 m/s
D. 30 m/s
Solution: After 1 s, v_y = 30 − 10 × 1 = 20 m/s while vₓ = 15 m/s stays constant; speed = √(15² + 20²) = 25 m/s.
Q15 — Projectile Motion · medium · numerical
A drone flying horizontally at 24 m/s at a height of 20 m releases a package. How far ahead of the release point, horizontally, does the package land? (g = 10 m/s²)
A. 40 m
B. 24 m
C. 96 m
D. 48 m ✓ Correct
Solution: Fall time: 20 = 5t² gives t = 2 s; the package keeps the drone speed, so horizontal distance = 24 × 2 = 48 m.
Q16 — Projectile Motion · medium · numerical
A sphere is projected so that its horizontal range is 4√3 times its maximum height. What is the angle of projection?
A. 37°
B. 45°
C. 30° ✓ Correct
D. 60°
Solution: H/R = tan θ / 4, so tan θ = 4H/R = 4/(4√3) = 1/√3, giving θ = 30°.
Q17 — Projectile Motion · medium · numerical
A ball is projected with a vertical velocity component of 25 m/s. At what two times after launch is it at a height of 20 m? (g = 10 m/s²)
A. 1 s and 5 s
B. 2 s and 3 s
C. 1 s and 4 s ✓ Correct
D. 2 s and 4 s
Solution: 20 = 25t − 5t² gives t² − 5t + 4 = 0, i.e. (t − 1)(t − 4) = 0, so t = 1 s on the way up and t = 4 s on the way down.
Q18 — Projectile Motion · medium · numerical
A sphere is thrown horizontally at 20 m/s from a tall tower. After how much time does its velocity point at 45° below the horizontal? (g = 10 m/s²)
A. 2√2 ≈ 2.8 s
B. 1 s
C. 2 s ✓ Correct
D. 4 s
Solution: At 45° the components are equal: v_y = gt = 10t must equal 20 m/s, so t = 2 s.
Q19 — Projectile Motion · medium · numerical
A ball thrown at 45° above the horizontal reaches a maximum height of 40 m. What was its launch speed? (g = 10 m/s²)
A. 80 m/s
B. 20 m/s
C. 20√2 ≈ 28.3 m/s
D. 40 m/s ✓ Correct
Solution: H = u² sin²45° / 2g = u²/40, so u² = 40 × 40 = 1600 and u = 40 m/s.
Q20 — Projectile Motion · medium · numerical
A trolley-mounted cannon fires a ball at 25 m/s at 53° above the horizontal (sin 53° = 0.8). What is its horizontal range on level ground? (g = 10 m/s²)
A. 45 m
B. 80 m
C. 60 m ✓ Correct
D. 75 m
Solution: Components: 25 cos 53° = 15 m/s and 25 sin 53° = 20 m/s; T = 2 × 20 / 10 = 4 s, so R = 15 × 4 = 60 m.