From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ____________ s.
Answer (integer)
3
Solution
<p>Based on the situation</p>
<p>$h = - u{t_1} + {1 \over 2}gt_1^2$ $\to$ throwing up ....... (i)</p>
<p>$h = u{t_2} + {1 \over 2}gt_2^2$ $\to$ throwing up ....... (ii)</p>
<p>$h = {1 \over 2}g{t^2}$ $\to$ dropping .......... (iii)</p>
<p>and $0 = u({t_1} - {t_2}) - {1 \over 2}g{({t_1} - {t_2})^2}$ ....... (iv)</p>
<p>solving above equations</p>
<p>$t = \sqrt {{t_1}{t_2}}$</p>
<p>$\Rightarrow t = \sqrt {6 \times 1.5} = 3\,s$</p>
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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