# 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

Ankit Sharma
30 Points
10 years ago

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

Jatin
26 Points
5 years ago
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
15 Points
4 years ago
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
ankit singh
askIITians Faculty 614 Points
2 years ago
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