Question bank/JEE Main/Physics

A man throws a packet from the top of the tower directly aiming
at his friend who is standing on the ground at a certain distance from
the base of the tower and this distance is same as the height of the
tower. If packet is thrown with a speed of 4 m /s and it hits the ground
midway between the tower base and his friend then height of the tower
(in m) is 0.4k. Find k. $(g = 10m/s^{2})$

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$\frac{4}{\sqrt{2}} \cdot t = \frac{h}{2}$....(1)
&$h = (\frac{4}{\sqrt{2}})t + \frac{1}{2}{gt}^{2}$....(2)
$\frac{h}{2} = \frac{1}{2}gt^{2}$
from (1) & (2)

Source: Grass-G / P-18 Physics paper 27 Nov Third.docx