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 28.6 m/s A 1.0-kg stone is thrown from the basket with an initial velocity of 15.9 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 5.70 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 28.6 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

Velocity of stone is, V2x = 11.6 m/sec, which is normal to V1y

speed of the basket, V2y = V1y = -28.6 m/sec

time for being falling down,tf = 5.75 sec

So,-h = V2y*t-4,9t2 = -28.6*5.7-(4.9)*5.72 = -326.4 m

So, h = 326 m over the ground

b)

hb = h+Vb*tf = 326-28.6*5.7 = 162.98 m

c)same as b answer.

hb = 162.98 m

D. We should find the horizontal and vertical velocity components of velocity as measured by an observer at rest in the basket.

Vbx = V2x = 15.9 m/sec

Vby = g*t = -9.8*5.7 = -55.86 m/sec (as basket was still )

E. we should find the horizontal and vertical velocity components before the rock hits the ground as measured by an observer at rest on the ground.

Vgx = Vbx = 15.9 m/sec

Vgy = Vby+V1y = -55.86-28.6 = -84.46 m/sec

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