A ball is projected vertically upwards from ground. It reaches a height ' h ' in time t₁, continues its motion and then takes a time t₂ to reach ground. The height h in terms of g, t₁ and t₂ is (g = acceleration due to gravity) (a) 1/2(gt₁)/(t₂) (b) 1/2gt₁t₂ (c) gt₁t₂ (d) 2gt₁t₂

Answer: (b). h = ut - 1/2gt² has two roots: going up at t₁, coming down at t₁'. Total flight time is t₁ + t₂; by symmetry the ball is at h again at t₁' = (t₁ + t₂) - t₁ = t₂. Product of roots of 1/2gt² - ut + h = 0: t₁ t₂ = 2h/g h = 1/2gt₁t₂. So the answer is (b).

A ball is projected vertically upwards from ground. It reaches a… | MHT CET 2023 (12 May, Shift 2)

Q17MHT CET 2023 · 12 MAY · SHIFT 2Medium

Motion Under Gravity

A ball is projected vertically upwards from ground. It reaches a height ' $h$ ' in time $t_1$ , continues its motion and then takes a time $t_2$ to reach ground. The height $h$ in terms of $g, t_1$ and $t_2$ is $(g =$ acceleration due to gravity)

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Answer: (b)

$h = ut - \frac{1}{2}gt^2$ has two roots: going up at $t_1$ , coming down at $t_1'$ .
Total flight time is $t_1 + t_2$ ; by symmetry the ball is at $h$ again at $t_1' = (t_1 + t_2) - t_1 = t_2$ .
Product of roots of $\frac{1}{2}gt^2 - ut + h = 0$ : $t_1 t_2 = \frac{2h}{g}$
$h = \frac{1}{2}gt_1t_2$ . So the answer is (b).

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Motion Under Gravity question from MHT CET 2023 (12 May, Shift 2)

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