Question

A man throws a packet from a tower directly aiming at his friend who is standing at a certain distance from the base which is same as a height of the tower. If packet is thrown with a speed of $4m/s$ and it hits the ground midway between the tower base & his friend. Find the height of the tower. (Take $g=9.8m/s^2$)

• A. $3.8$
• B. $4.4$
• C. $3.2$
• D. $2.2$

Answer 1

Answer: (C)

$3.2$

$\frac{4}{\sqrt{2}}.t=\frac{h}{2}...\left(i\right)$
& $h=\left(\frac{4}{\sqrt{2}}\right)t+\frac{1}{2}g{t}^2...\left(ii\right)$

from (i) & (ii)
$\frac{h}{2}=\frac{1}{2}g\frac{4^2}{2}$

$\Rightarrow \frac{h}{2}=\frac{1}{2}g{\left(\frac{h}{4\sqrt{2}}\right)}^2$

$h=\frac{16\times 2}{10}=3.2m$