Question

A body is projected horizontally with a speed of 20 m/s from the top of a very high lower. What will be its speed after nearly 5 see? $(g=10m/s^2)$

Answer 1

Given,

Horizontal velocity, $u_x=20m{s}^{-1}$

Apply kinematic equation of motion

$v_y=u_y+a_{y}t$

$v_y=0+10\times 5=50m{s}^{-1}$

Resultant velocity, $V_R=\sqrt{v_y^2+u_x^2}=\sqrt{{20}^2+{50}^2}=10\sqrt{29}m{s}^{-1}$

Hence, resultant velocity after 5 sec $10\sqrt{29}m{s}^{-1}$