A 124-kg balloon carrying a 22-kg basket is descending with a constant downward

Question

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 12.0 m/s A 1.0-kg stone is thrown from the basket with an initial velocity of 13.6 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.90 S after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 12.0 m/S. How high was the balloon when the rock was thrown out? How high is the balloon when the rock hits the ground? At the instant the rock hits the ground, how far is it from the basket? Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest in the basket. Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest on the ground.

Explanation / Answer

For much of the problem you can act like the object was thrown down with a vertical velocity fo 12 m/s and a horizontal velocity of 13.6 m/s. I am defining the down direction as positive, to make things easier. So gravity is positive and the balloon descends in a positive y direction.

a) u = 13.2 m/s
a = 9.81 m/s^2
t = 6.90 s

Thats all you need, three variable

s = vt + 1/2 a t^2

s = 12 m/s * 6.90 s + 1/2 * (9.81 m/s^2) * (6.90 s)^2

s =316.32 m

b)The balloon was 316.32 m high when the stone was thrown, but it was moving down at 12 m/s for 6.90 seconds that the stone took to hit the ground.

height = 316.32 m - 12 m/s * 6.90 sec

height = 233.52 m

c) we just figured out the VERTICAL distance from the balloon to the ground, now we need the horizontal distance

Horizontal velocity is constant so X = 6.90 sec * 13.6 m/s = 93.84 m

Now you must use the pythagean theorem with the horizontal and vertical to get the total distance.

d)The horizontal component is constant and observed the same by the guy in the basket, 13.6m/s

The vertical component as seen from the ground is v = u + at

v = 12 m/s + 9.81 m/s * 6.90 sec

v = 79.68 m/s

That is the answer to e) actually,

for what the observer on the balloon sees, he sees the same horizontal velocity, but u less the speed observed on the ground

Vertical velocity = 9.81 * 6.90 sec = 67.68m/s

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