Question

A bomb is released from an aeroplane which is moving horizontally with a velocity $196\text{m/s}$ at a height of $980\text{m}$ from the ground. Find the velocity with which the bomb hits the ground.

• A. $196\text{m/s}$
• B. $138.5\text{m/s}$
• C. $240\text{m/s}$
• D. $144\text{m/s}$

Answer 1

The correct option is C $240\text{m/s}$

Given, vertical velocity of aeroplane, $u_y=0$
& vertical acceleration $a_y=g$
So, using 2nd eqn of motion: (take downward +ve)
$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$980=0+\frac{1}{2}\times 9.8\times t^2$
$t^2=200$ or $t=14.14\text{s}$

Velocity in y - direction when bomb hit the ground
$v_y=u_y+a_{y}t$
$=0+9.8\times 14.14$
$=138.5\text{m/s}$
Velocity of bomb in x - direction when it hits the ground,
$v_x=196\text{m/s}$

So, velocity of bomb when it hits the ground
$v=\sqrt{v_x^2+v_y^2}=\sqrt{(196{)}^2+(138.5{)}^2}$
$\approx \sqrt{57600}$
$=240\text{m/s}$