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 20.0 m/s. A 1.0-kg stone is thrown from the basket with an initial velocity of 15.0 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 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. (a) How high is the balloon when the rock is thrown? (b) How high is the ballon when the rock hits the ground? (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and veritcal velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground.

Explanation / Answer

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 = 20 m/s
a = 9.81 m/s^2
t = 5.0 s
then
s = vt + 1/2 a t^2

s = 20 m/s * 5.0 s + 1/2 * (9.81 m/s^2) * (5.0 s)^2

s = 222.625 m

b)The balloon was 222.625 m high when the stone was thrown, but it was moving down at 20 m/s for 5.0 seconds that the stone took to hit the ground.

height = 222.625 m - 20.0 m/s * 5.0 sec

height = 122.625 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 = 5.0 sec * 15.0 m/s = 75 m

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

t.D = sqrt(122.652 + 752)

total distance = 143.76m

d)The horizontal component is constant and observed the same by the guy in the basket, 15 m/s

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

v = 20 m/s + 9.81 m/s * 5.0 sec

v = 69.05 m/s

(ii)

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 * 5.0 sec = 49.05 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.