Question

A hot air balloon moves vertically upwards at a constant velocity of 1.5 m s−1. A person standing on the ground below throws a ball towards the balloon with vertical velocity of 10 m s−1. The ball does not reach the balloon.

Calculate the time at which the ball comes closest to the balloon after it is thrown.

(You may ignore air resistance. Take the acceleration due to gravity to be 10 m s−2.)

Answer 1

Answer:

1.7 sec

Explanation:

S1= 1.5(t)

S2=10t - 1/2 × 10 (t)^2

the ball is closest to the balloon that means approximately s1=s2

or,1.5 t= 10t- 5t^2

or, t= 8.5/5=1.7 sec

HOPE IT HELPS????!?