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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

8 years ago

Answers : (4)

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

8 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
						
3 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
2 years ago
ankit singh
askIITians Faculty
596 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 thank u ankit singh
v=5
so his velocity on reaching the ground is 5m/s
3 months ago
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