Question

A plane flying horizontally at $100\text{m/s}$ releases an object which reaches the ground in $10\text{s}$. What will be the angle from horizontal, at which it hits the ground? Take $g=10{\text{m/s}}^2$.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

Object is dropped from plane, hence it's horizontal component of velocity $u_x$ will remain constant and will be equal to the horizontal velocity of plane.
Horizontal velocity of plane$=100\text{m/s}$
$\Rightarrow v_x=u_x=100\text{m/s}$
$\because$ Object reaches ground in $10\text{s}$
$v_y=u_y+a_{y}t$, taking vertically downwards as $+vey$ direction
$\therefore v_y=0+g\times t=10\times 10=100\text{m/s}$
From vector triangle representing velocity while striking gound:

$\tan \theta =\frac{v_y}{v_x}=\frac{100}{100}$
$\Rightarrow \tan \theta =1$
$\therefore \theta ={45}^{\circ }$