A parachutist drops freely from an aeroplane for 10 s before the parachute opens out. Then he descends with a net retardation of 2. 5 m/s² . If he bails out of the plane at a height of 2495 m and g = 10 m/sec², his velocity on reaching the ground will be

The velocity v acquired by the parachutist after 10 sec is 
$\mathrm{v}=\mathrm{u}+\mathrm{gt}=0+10\times 10=100\mathrm{m}/\mathrm{s}$
Let s₁ be the height of fall for 10 sec. Then
${\mathrm{s}}_1=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2=0+\frac{1}{2}\times 10\times 100=500\mathrm{m}$.Then
distance travelled by the parachutist under retardation
s₂ = 2495 - 500 = 1995 metre
Let v' be his velocity on reaching the ground.
Then v'² -v² =-2as,
or v'² -(100)² =-2x2.5x1995
Solving, we get v' = 5 m/s