Question

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

• A. $30$
• B. $60$
• C. $120$
• D. $20$

Answer 1

The correct option is A $30$

Initial horizontal component of velocity $u_x=20m/s$

Since there is no acceleration in horizontal direction, so horizontal component of velocity remains the same.

Final horizontal component of velocity of particle when it reaches the ground $v_x=u_x=20m/s$

Vertical direction :

Final vertical component of velocity of particle when it reaches the ground $v_y=\sqrt{2gh}$

where $h=25m$ and $g=10m/s^2$

$\therefore$ $v_y=\sqrt{2\times 10\times 25}=\sqrt{500}m/s$

Net velocity of particle $v=\sqrt{(v_x{)}^2+(v_y{)}^2}=\sqrt{{20}^2+500}=\sqrt{900}=30m/s$