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 27.1 m/s . A 1.0-kg stone is thrown from the basket with an initial velocity of 19.6 m/sperpendicular 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 17.1 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 27.1 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

a) The height of the balloon when the rock was thrown out, H = voy*t + 0.5*g*t^2

= 0*t + 0.5*9.8*17.1^2

= 1432.8 m <<<<<<<<<<-------------Answer

b)The ballon also hits the ground at the same instant.

beacuse y-component of velcities for both are same.

c) distance travelled by rock in hrizontl direction before hitting the ground,x = vox*t

= 19.6*17.1

= 335.16

d) x-component of velocity with respect oborver in the ballon, vx = vox

= 19.6 m/s (because x-component of velocity remainse same)

but, y-component of velocity with respect oborver in the balloon, vy = 0 (since the ballon and stone have same velocity in y-direction)

e) x-component of velocity with respect oborver at the ground,vx = vox

= 19.6 m/s (because x-component of velocity remainse same)

y-component of velocity of stone with respect obsorver at the ground,vy = voy + g*t

= 0 + 9.8*17.1

= 167.58 m/s

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