Question

A relief aeroplane is flying at a constant height of $1960\text{m}$ with $600\text{km/hr}$ speed above the ground towards a point directly over a person struggling in flood water. At what angle of sight with the vertical should the pilot release a survival kit if it has to reach the person in water ?
$(g=9.8{\text{m/s}}^2)$

• A. ${30}^{\circ }$
• B. ${37}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D ${60}^{\circ }$


Since, the Projectile is being released from a plane moving horizontally. So

Initial Speed of projectile = horizontal speed of plane

So, initial speed of projectile,
$u_x=\frac{500}{3}\text{m/s}$ and $u_y=0$

Let's say it takes $t\text{second}$ to reach the man.

Now using second equation of motion,

$h=u_y.t+\frac{1}{2}g{t}^2$

$\Rightarrow 1960=(0\times t)+\frac{1}{2}\times 9.8\times t^2$

$\Rightarrow t=\sqrt{\frac{2\times 1960}{9.8}}=\sqrt{400}=20\text{s}$

In the same time it travels $x$ distance horizontally

$\therefore x=u_x\times t=\frac{500}{3}\times 20$

$\Rightarrow x=\frac{10000}{3}\text{m}$


From above diagram,

$\tan \theta =\frac{x}{h}=\frac{10000}{3\times 1960}=1.7=\sqrt{3}$

$\Rightarrow \theta ={60}^{\circ }$

Hence, option $\left(d\right)$ is correct.