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

Question

A 124- balloon carrying a 22- basket is descending with a constant downward velocity of 21.2 A 1.0- stone is thrown from the basket with an initial velocity of 10.4 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 5.90 after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 21.2 .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.D)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

From h = Vot + gt^2, where h = height, Vo = initial downward velocity, g = gravity = 9.8m/s and t = time in seconds
A)
h = 21.2(5.90) + (9.8(5.90^2))/2 = 295.649 m

(B) In the horizontal direction, initial velocity v = 21.2 m/s
Acceleration = 0

So distance traveled by the stone in the horizontal direction while it is in the air

(i.e. for 4 s) = v * t = 21.2 * 5.9 = 125.08 m

L = 125.08 m

(C) Consider the motion of the balloon after the stone is thrown : Velocity, v = 10.4 m/s
Time = 5.9 s
Distance, s = vt = 10.4*5.9 = 61.36 m

So height of the balloon above the ground after 5.9s = 295.08 - 61.36 = 234.44 m

Now distance between the basket and the balloon = sqrt (60^2 + 78^2) = 98.4 m (You will be able to understand this better if you make the figure of the balloon and the basket after 4s)

So d = 98.4 m

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