Question

A particle is projected up with a velocity of 20 m $s^{-1}$ from a tower of height 25 m. Its velocity on reaching the ground is _________ m $s^{-1}$.

• A. 20
• B. 40
• C. 30
• D. 10

Answer 1

Answer: (C)

30

Taking upward direction to be positive.

Initial velocity $u=20m/s$

Acceleration $a=-g=-10m/s^2$

Displacement $S=-25m$

Using $v^2-u^2=2aS$

Or $v^2-(20{)}^2=2(-10\left)\right(-25)$

$\therefore$ $v^2-400=500$

Or $v=\sqrt{900}=30m/s$