Question

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 10.8 m/s . A 1.0-kg stone is thrown from the basket with an initial velocity of 14.3 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 18.8 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 10.8 m/s.

(a) How high was the balloon when the rock was thrown out?
(b) How high is the balloon when the rock hits the ground?
(c) At the instant the rock hits the ground, how far is it from the basket?

Answer 1

Answer:

a) -1529m

b)-1326m

c)268.84 meters.

Explanation:

a) Since the stone is thrown horizontally, its initial vertical velocity is the same as the basket. Let’s use the following equation to determine the vertical distance it moves in 18.8 seconds.

According to kinematic equation the displacement is given by

$h_1 = v_{0y} y - \frac{1}{2}gt^2$

$h_1$=10.8 * 18.8 -  ½ * 9.8 * 18.8² => -1529m

The negative sign is due to the direction.

b)while the stone was travelling $h_1$ for 18.8s the balloon was also travelling the displacement $h_b$ with $v_o_y$. so $h_b$  is given by

$h_b=v_o_y t= - 10.8\times18.8$=> -203.04m

The height above the eart is given by,

$h_2=h_1-h_b = -1529+203.04 => -1326$m

c)At the instant the rock hits the ground, how far is it from the basket?  

This is the product of its initial horizontal velocity and the time.

d = 14.3 * 18.8 = 268.84 meters.