Question

A parachutist drops first freely from an aeroplane for 10s and then parachute opens out. Now he descends with a net retardation of 2.5 m/ $s^2$. If he falls out of the plane at a height of 2495 m and g=10 m/$s^2$, his velocity on reaching the ground will be

• A. 5 m/s
• B. 10 m/s
• C. 15 m/s
• D. 20 m/s

Answer 1

The correct option is A

5 m/s

The velocity v acquired by the parachutist after 10 s.

V = u + gt = 0 + 10 $\times$ 10 = 100 ms

Then ,

$s_1$ = ut + $\frac{1}{2}$ g $t^2$ = 0 + $\frac{1}{2}$ $\times$ 10 $\times$ ${10}^2$ = 500 m

The distance travelled by the parachutist under retardation, $s_2$ = 2495 - 500 = 1995 m.

Then

${v_g}^2$ - $v^2$ = 2a $s_2$

or ${v_g}^2$ - ${\left(100\right)}^2$ = 2 $\times$ (- 2.5) $\times$ 1995

or $v_g$ = 5 m/s