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Grade: 11

                        

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

2 years ago

Answers : (1)

Arun
24742 Points
							
 

The initial velocity of the parachutist is zero

 

now V=u+gt

 

V=0+(-10)10

 

V=-100 (-sign as the velocity is in downward direction)

 

Now S=ut+1/2ut^2

 

so S=0+1/2(-10)*100

 

S=500

 

Now H=2495-500

 

=1995

 

NOw v^2=u^2+2ah

 

v^2=10000-2*2.5*1995

 

v^2=25

 

v=5

 

so his velocity on reaching the ground is 5m/s

2 years ago
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