Question

A body thrown up with a velocity of 98 m/s reaches a point P in its path 7 second after projection. Since its projection it comes back to the same position after:

• A. 13s
• B. 14s
• C. 6s
• D. 22s

Answer 1

The correct option is A 13s

Here, $u=98m/s$, $t_1=7s$. When body reaches maximum height, its velocity becomes $v_1=0$.
Maximum height reached can be obtained using:
$v^2=u^2+2ah$
$h_{max}=\frac{u^2}{2g}$
Now considering initial velocity as $v_1=0$, displacement as $s=\frac{u^2}{2g}$,

We can find time for the body to come back from the maximum height using:
$s=ut+\frac{1}{2}a{t}^2\frac{u^2}{2g}=\frac{1}{2}g{t}^{2}t=\frac{u}{g}$
Thus, total time, $t=\frac{2u}{g}=20s$
Now the time it takes to reach point P from ground and to reach from point P to ground is same i.e. $7s$ each time.

Thus, it takes $20s-7s=13s$ to come back to the same point P since its projection.