Answer:

Given:

The height of point A $H_1=20m$

The height of point A $H_2=16m$

Horizontal distance between A and B is $d=40m$

To find:

Minimum horizontal velocity $“u(m/s)”$ so that the rock just clears the obstruction B

Solution:

• Let the time be$t=0s$ at the point where the boy throws the stone horizontally at point A
• Let the time be$t=Ts$ at the point where the stone reaches B
• Considering the vertical travel of the stone
• We know that for free falling body with zero initial velocity$s= {1\over2}gt^2$
• Where s is vertical displacement$(m)$

$t$ is time $(s)$ $g$ is acceleration due to gravity $(m/s^2)$ $s=H_1-H_2=20-16=4m$ * Substituting these values $4={1\over2}\times9.81\times T^2$ $T= 0.903s$ * So the stone has to travel the horizontal distance d in time T to satisfy the given condition * So we get the initial horizontal velocity$u={d\over t}={40\over0.903}= 44.29 m/s$ * So the minimum horizontal velocity$ “u(m/s)”$ so that the rock just clears the obstruction B is$44.29m/s$