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Grade: 11
        a parachutist first drops freely from an aeroplane for 10 seconds and then his parachute opens out.now he decends with a retardation of 2.5 m/s2.if he bails out of plane at an height of 2495m .find his velocity on reaching the ground
7 years ago

Answers : (3)

Ankit Sharma
30 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

7 years ago
Jatin
26 Points
							The initial velocity of the parachutist is zeronow V=u+gtV=0+(-10)10V=-100 (-sign as the velocity is in downward direction)Now S=ut+1/2ut^2so S=0+1/2(-10)*100S=500Now H=2495-500=1995NOw v^2=u^2+2ahv^2=10000-2*2.5*1995v^2=25v=5so his velocity on reaching the ground is 5m/s
						
2 years ago
Hammad khan
15 Points
							
The velocity v acquired by the parachutist after 10s : 
V=u+gt=0+1/2×10+×10^2=100
Then, s1=ut+1/2gt^2=0+1/2×10×10^2=500
The distance travelled the parachutist under retardation is
S2=2495-500=1995
Let vg  by the velocity on reaching the ground. Then
Vg^2-v^2=2as2
Or.  Vg^2-(100)^2=2×(-2.5)×1995 or vg= 5m/s
one year ago
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