Question

A parachutist drops freely from an aeroplane for 5 sec. Then he opens the parachute and descends with a retardation of 8
m/s2 and strikes the ground with a velocity of 30 m/s. The height, at which he drops out of the plane, is (g = 10 m/s2)​

Answer 1

Answer:

5m/s

Explanation:

nitial velocity (u)=0

From t=0 to t=10 free fall

V=0+(10)(10)

V=100m.s

Distance covered=+  

2

1

​  

×10×10  

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)(1995)

=10000−9975

=25

⇒v=5m/s