Question

An airplane flying $490\text{m}$ above ground level at $100\text{m/s}$, releases a bomb. The horizontal distance it has to travel before it hits the ground is (Take $g=9.8{\text{m/s}}^2$)

• A. $0.1\text{km}$
• B. $1\text{km}$
• C. $2\text{km}$
• D. $2.5\text{km}$

Answer 1

The correct option is B $1\text{km}$

Finding the time required to fall to the ground,
Let downward direction be positive,
$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$h=\frac{1}{2}{a}_y{t}^2$ (since y-component of initial velocity is zero)
$h=\frac{1}{2}g{t}^2$
$t=\sqrt{\frac{2h}{g}}$

Horizontal distance it has to travel is given by
$x=u_{x}t$ = $u_x\sqrt{\frac{2h}{g}}=100\times \sqrt{\frac{2\times 490}{9.8}}=1000\text{m}=1\text{km}$