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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
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
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
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
The velocity v acquired by the parachutist after 10s : V=u+gt=0+1/2×10+×10^2=100Then, s1=ut+1/2gt^2=0+1/2×10×10^2=500The distance travelled the parachutist under retardation isS2=2495-500=1995Let vg by the velocity on reaching the ground. ThenVg^2-v^2=2as2Or. Vg^2-(100)^2=2×(-2.5)×1995 or vg= 5m/s
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