Q10. A projectile at 60° reaches maximum height 15 m. If g = 10 m/s², what was launch speed?
[SSC Scientific Assistant Physics — 24 Nov 2017, Shift 1]
(A) 10 m/s
(B) 10√3 m/s
(C) 20 m/s
(D) 20√3 m/s
Answer: (C) 20 m/s
Explanation: The maximum height is H = u²sin²θ/(2g). Substituting H = 15, g = 10 and sin²60° = 3/4 gives 15 = 3u²/80, so u² = 400 and u = 20 m/s.
Q9. A ball launched at 30 m/s, 15° above horizontal, lands at the same level. Take g = 10 m/s². Find range.
[SSC Scientific Assistant Physics — 24 Nov 2017, Shift 2]
(A) 40 m
(B) 50 m
(C) 45 m
(D) 35 m
Answer: (C) 45 m
Explanation: Horizontal range on level ground is R = u²sin2θ/g. Here u² = 900 and sin30° = ½, giving R = 900 × ½/10 = 45 m under ideal projectile assumptions.
Q8. Which expression gives the maximum height of a projectile launched at speed v and angle θ?
[SSC Scientific Assistant Physics — 23 Nov 2017, Shift 2]
(A) v sin²θ/(2g)
(B) v²sin²θ/(2g)
(C) v²sin²θ/g
(D) v sin²θ/g
Answer: (B) v²sin²θ/(2g)
Explanation: At maximum height the vertical component of velocity is zero. The launch vertical speed is v sinθ. Using 0 = (v sinθ)² − 2gH yields H = v²sin²θ/(2g).
Q7. A projectile begins at speed v and elevation angle θ. Which formula gives time to reach greatest height?
[SSC Scientific Assistant Physics — 22 Nov 2017, Shift 2]
(A) v sinθ/g
(B) v²sinθ/g
(C) v²sin²θ/g
(D) v sin²θ/g
Answer: (A) v sinθ/g
Explanation: Vertical velocity changes according to v_y = v sinθ − gt. At maximum height, v_y = 0. Solving gives t = v sinθ/g. The total flight time back to the launch level is twice this time.
Q6. Which quantity remains constant when a ball is thrown vertically upward, neglecting air resistance?
[SSC CGL — 12 Aug 2017, Shift 2]
(A) Total mechanical energy
(B) Displacement
(C) Velocity
(D) Kinetic energy
Answer: (A) Total mechanical energy
Explanation: As the ball rises, kinetic energy decreases while gravitational potential energy increases. The sum of those two energies stays constant if no dissipative force acts. Velocity and displacement change with time, so only total mechanical energy is conserved.
Q5. A ball is projected vertically at 25.2 m/s. When does it reach maximum height? Take g = 9.8 m/s².
[UPSC NDA-II GAT — 18 Sep 2016]
(A) 5.14 s
(B) 3.57 s
(C) 2.57 s
(D) 1.29 s
Answer: (C) 2.57 s
Explanation: The upward velocity decreases by g every second. At the top v = 0, so t = u/g = 25.2/9.8 ≈ 2.57 s. This is the ascent time, not the complete return time.
Q4. One ball rises from ground at 20 m/s; another falls from 40 m at 20 m/s. When and where do they meet? g = 9.8 m/s².
[UPSC NDA-II GAT — 18 Sep 2016]
(A) 3 s; 30.2 m high
(B) 2 s; 20.1 m high
(C) 1 s; 15.1 m high
(D) 5 s; 20 m high
Answer: (C) 1 s; 15.1 m high
Explanation: Both balls have the same downward acceleration, so relative acceleration is zero. Initial closing speed is 40 m/s across a 40 m separation; they meet after 1 s. The rising ball reaches 20 − 4.9 = 15.1 m.
Q3. A body falls from 20 m. After descending 5 m, what portion of its initial gravitational potential energy is lost?
[UPSC NDA-I GAT — 17 Apr 2016]
(A) One-quarter of total energy
(B) One-quarter of initial potential energy
(C) One-quarter of potential energy is gained
(D) Three-quarters of total energy is gained
Answer: (B) One-quarter of initial potential energy
Explanation: The initial potential energy relative to ground is mg × 20. After falling 5 m the height is 15 m, so remaining potential energy is mg × 15. The loss is 5/20, or one-quarter, of the initial value.
Q2. What happens to a ball’s weight when it is carried from Earth’s equator to a pole?
[WBCS Preliminary Examination — 2015]
(A) Its mass increases
(B) Mass and weight both change
(C) Its weight increases
(D) Its weight decreases
Answer: (C) Its weight increases
Explanation: Earth is slightly flattened at the poles and its rotation reduces effective gravity at the equator. The acceleration due to gravity is thus higher at poles. With mass unchanged, weight W = mg increases as the ball moves poleward.
Q1. Why does an artificial Earth satellite remain in orbit instead of falling directly to the surface?
[UPSC Civil Services Preliminary Examination — 2011]
(A) Earth’s attraction vanishes at orbital altitude
(B) The Moon cancels Earth’s attraction
(C) Gravity supplies its steady speed
(D) Earth’s gravity supplies the necessary orbital acceleration
Answer: (D) Earth’s gravity supplies the necessary orbital acceleration
Explanation: Earth’s gravity continuously accelerates the satellite toward Earth and bends its path into an orbit. The satellite also has sufficient tangential speed to keep missing the surface. It is actually in continuous free fall, not beyond gravitational influence.
