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

Answer:

y=39.057 m

Explanation:

Using Kinematic relation

$s=ut+ \frac{1}{2}at^2$

given u= 5m/s

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

$s=5t- \frac{1}{2}(9.80)t^2$

Also, we know that

v=u+at

v=5-9.80t

at time t= 0.250 sec

$s=5\times0.25- \frac{1}{2}(9.80)0.25^2$

s=0.94375 m

now position of sandbag

y= 40-0.94375

y=39.057 m