Question

A plane is flying horizontal at $98m{s}^{-1}$ and released an object which reaches the ground in $10s$. The angle made by it while hitting the ground is:

• A. ${55}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${75}^o$

Answer 1

The correct option is B ${45}^o$

The angle can be find just by finding vertical component and horizontal component of velocity with which it will hit the ground.

So, considering for vertical motion , velocity after 10 sec will be

$v=0+gt$ (As, initially downward component of velocity was zero)

So, $v=9.8\times 10=98m/s$

Now, horizontal component of velocity remains constant through out the motion i.e. $98m/s$ (As because this velocity was imparted to the object while releasing from the plane moving with this amount of velocity)

So, angle made with the ground while hitting $={\tan }^{-1}\left(\frac{98}{98}\right)={45}^o$