Question bank/JEE Main/Physics

A ball is thrown vertically upwards from the ground. It crosses a
point at the height of 25 m. twice at an interval of 4 secs. The ball
was thrown with the velocity of

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Let initial velocity of the ball = u
$\therefore 25 = ut - \frac{1}{2}gt^{2} = ut - 5t^{2}$
$\Rightarrow 5t^{2} - ut + 25 = 0$
If t_1and t_2 are the two solutions,
[IMAGE] Then, $t_{2} - t_{1} = \sqrt{{(t_{2} + t_{1})}^{2} - 4t_{1}t_{2}}$
$\Rightarrow 4^{2} = \frac{u^{2}}{25} - 4 \times \frac{25}{5}$
$\Rightarrow 16 = \frac{u^{2}}{25} - 20 \Rightarrow u^{2} = 25 \times 36$
$\Rightarrow u = 30m/s$

Source: Grass-G / P-16 Physics paper 25 Nov First.docx