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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
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B
10ms1
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C
15ms1
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D
20ms1
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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

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