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 13.7 m/s . A 1.0-kg stone is thrown from the basket with an initial velocity of 18.7 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 13.3 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 13.7 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?

D. Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest in the basket.

E. 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)

Consider the vertical motion of the rock :

Voy = initial velocity = -13.7 m/s                     since rock has same downward velocity as ballon

ay = acceleration = - 9.8 m/2

t = time taken to reach the ground = 13.3 s

Y = vertical displacement = height from which the rock was thrown

Using the equation

Y = Voy t + (0.5) ay t2

Y = (-13.7) (13.3) + (0.5) (-9.8) (13.3)2

Y = - 1048.67 m

so height = 1048.67 m

B)

Distance travelled by the ballon in down direction in 13.3 sec = 13.7 x 13.3 = 182.21 m

height of balloon = 1048.67 - 182.21 = 866.46 m

C)

Along the horizontal direction , Vox = 18.7 m/s

Distance from balloon in horizontal direction = X = Vox t = 18.7 x 13.3 = 248.71 m

Distance from balloon in horizontal direction = Y = 866.46 m

distance = sqrt(X2 + Y2) = sqrt(248.712 + 866.462) = 901.45 m

D)

Horizontal component = 18.7

Vertical component = at = 9.8 x 13.3 = 130.34

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