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2nd Equation of Motion

A ballon is r...

Question

A ballon is rising vertically upwards at a velocity of 3m/s. WHen it is at a height of 45m from the ground, a parachutist bails out from it. After 3s he opens his parachute and decelerates at constant rate of 5m/s. What was the height of parachutist above the ground when he opened his parachute? Tage g = 10m/s^-2.

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Solution

When the parachutist bails out, he has same velocity of the balloon.
So,
u=3m/s in upward direction.
g= -10ms2
t=3sec
so by equation of motion:
s=ut+1/at²
=3x3+1/2(-10)x 3x3
= 9-45s
= -36m
Here negative sign indicates that it is directed downwards.
So height from the ground when he opened parachute is =45-36=9m Ans.

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