Question

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

• A. $5m{s}^{-1}$
• B. $10m{s}^{-1}$
• C. $15m{s}^{-1}$
• D. $20m{s}^{-1}$

Answer 1

Answer: (A)

$5m{s}^{-1}$

Initial velocity $\left(u\right)=0$

From $t=0$ to $t=10$ free fall

$V=0+\left(10\right)\left(10\right)V=100m.s$

Distance covered$=+\frac{1}{2}\times 10\times {10}^2=500m$

After this retardation of $a=2.5m/s^2$

Now from here tall it reaches ground-

Distance covered $=$ total distance$-$the distance covered in first $10sec$

$=(2495-500)m=1995m$

Now, By $3^{rd}$ equation of motion-

$v^2=u^2+2as{v}^2=(100{)}^2+2(-2.5\left)\right(1995)=10000-9975=25\Rightarrow v=5m/s$