A plane is flying horizontally at $98m/s$ and releases an object which reaches the ground in $10$ sec. The angle made by it while hitting the ground is:

The correct option is B $45$
The correct answer is B.

Given,

$v=98m/s,t=10s$

The vertical component and the horizontal component of velocity with it will hit the ground.

so, considering for vertical motion, velocity 10s

$v=0+gt$

so,

$v=9.8\times 10$

$=98m/s$

so,

The angle made with the ground while hitting is $ta{n}^{-1}\left(\frac{98}{98}\right)$

=${45}^0$