Question

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

A
5ms1
B
10ms1
C
15ms1
D
20ms1

Solution

## The correct option is D $$5 ms^{-1}$$Initial velocity $$(u)=0$$From $$t=0$$ to $$t=10$$ free fall$$V=0+(10)(10)\\V=100m.s$$Distance covered$$=+\cfrac{1}{2}\times10\times10^2\\=500m$$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 $$10 sec$$$$=(2495-500)m\\=1995m$$Now, By $$3^{rd}$$ equation of motion-$$v^2=u^2+2as\\v^2=(100)^2+2(-2.5)(1995)\\=10000-9975\\=25\\ \Rightarrow v=5m/s$$Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More