Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground. After the sandbag is released, it is in free fall. Find the position of the sandbag at 0.250 s after its release.

Answer 1

y=39.057 m

Step-by-step explanation:

Using Kinematic relation

`s=ut+ (1)/(2)at^2`

given u= 5m/s

a=g= -9.81
`m/s^2`( directed downward)

`s=5t- (1)/(2)(9.80)t^2`

Also, we know that

v=u+at

v=5-9.80t

at time t= 0.250 sec

`s=5*0.25- (1)/(2)(9.80)0.25^2`

s=0.94375 m

now position of sandbag

y= 40-0.94375

y=39.057 m