A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of
- (a)8 m/s
- (b)12 m/s
- (c)16 m/s
- (d)20 m/s
Correct — D, 20 m/s. At the highest point the ball's velocity is momentarily zero. Using the equation of motion v² = u² − 2gh with v = 0 at the top gives u = √(2gh). Taking g ≈ 10 m/s² and h = 20 m, u = √(2 × 10 × 20) = √400 = 20 m/s. (Using g = 9.8 m/s² gives about 19.8 m/s, which rounds to the same 'approximately 20 m/s'.)
- (a)8 m/s — An initial speed of 8 m/s reaches only h = u²/2g = 64/20 ≈ 3.2 m, far short of 20 m.
- (b)12 m/s — 12 m/s reaches only about 144/20 ≈ 7.2 m, well below the required 20 m.
- (c)16 m/s — 16 m/s reaches about 256/20 ≈ 12.8 m; it still falls short of the 20 m maximum height.
For a body thrown straight up, gravity decelerates it until its velocity becomes zero at the maximum height. The kinematic relation v² = u² − 2gh (with the upward direction positive and g the acceleration due to gravity) links the launch speed to the height reached.
Set v = 0 at the top and solve for the launch speed u = √(2gh). The same result follows from energy conservation: the initial kinetic energy ½mu² converts fully into potential energy mgh, giving u = √(2gh). The distractors are the speeds that would produce smaller heights.
- At the maximum height of vertical projectile motion the velocity is zero.
- v² = u² − 2gh is the relevant equation of motion under gravity.
- Launch speed u = √(2gh); with g ≈ 10 m/s² and h = 20 m, u = 20 m/s.
- The result is independent of the ball's mass.
- Forgetting that the velocity is zero at the top and mis-setting the equation.
- Expecting the answer to depend on the ball's mass — it does not.
Asked as a one-step kinematics problem: given the maximum height, back-calculate the launch speed using v² = u² − 2gh.
No directly related past PYQ was found.
- practice — not a real PYQ
A stone is thrown vertically upward with a speed of 10 m/s. Taking g = 10 m/s², the maximum height it reaches is
- (a)2.5 m
- (b)5 m
- (c)10 m
- (d)20 m
Answer(b) 5 m — h = u²/2g = 100/20 = 5 m.
- practice — not a real PYQ
At the highest point of the vertical upward motion of a ball, its
- (a)velocity is maximum
- (b)velocity is zero and acceleration is g downward
- (c)acceleration is zero
- (d)velocity and acceleration are both zero
Answer(b) velocity is zero and acceleration is g downward — gravity still acts even when the speed is momentarily zero.