Question

Question No. 25
The velocity (m/s) of a parachutist is given as a function of time (seconds) by 55
seconds, the distance traveled by the body from t = 2 to 1 =12 seconds is estimated most
O 341.43 m
O 429.05 m
O 428.97 m
O 703.50 m​

Answer 1

Answer:

Initial velocity (u)=0

From t=0 to t=10 free fall

V=0+(10)(10)

V=100m.s

Distance covered=+

2

1

×10×10

2

=500m

After this retardation of a=2.5m/s

2

Now from here tall it reaches ground-

Distance covered = total distance−the distance covered in first 10sec

=(2495−500)m

=1995m

Now, By 3^{rd} equation of motion-

v

2

=u

2

+2as

v

2

=(100)

2

+2(−2.5)(1995)

=10000−9975

=25

⇒v=5m/s