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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

siddharth ojha , 13 Years ago

Grade 11

4 Answers

Ankit Sharma

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

Last Activity: 13 Years ago

Jatin

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

Last Activity: 8 Years ago

Hammad khan

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

Last Activity: 7 Years ago

ankit singh

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

Last Activity: 5 Years ago

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