Question

A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is vo = 1.55 m/s. Knowing that at the time of the release the balloon was 85.8 m above the ground, determine the time, T, it takes the bag to reach the ground from the moment of its release.

Answer 1

Answer:

t = 4.35 s

Explanation:

Since the balloon is moving upwards while the sand bag is dropped from it

so here the velocity of sand bag is same as the velocity of balloon

so here we can use kinematics to find the time it will take to reach the ground

$\Delta y = v_y t + \frac{1}{2} gt^2$

here we know that since sand bag is dropped down so we have

$\Delta y = -85.8 m$

initial upward speed is

$v_y = 1.55 m/s$

also we know that gravity is downwards so we have

$a = - 9.8 m/s^2$

so here we have

$-85.8 = 1.55 t - \frac{1}{2}(9.8) t^2$

$4.9 t^2 - 1.55 t - 85.8 = 0$

$t = 4.35 s$