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 .

At the instant the rock hits the ground, how far is it from the basket?

Explanation / Answer

Here ,

time of flight of stone , tf = 17.1 s

Using second equation of motion

h = u*t + 0.5 * at^2

h = 27.1 * 17.1 + 0.5 * 9.8 * 17.1^2

h = 1896.22 m

Now, horizontal distance travelled by stone , x = 19.6 * 17.1

x = 335.2 m

vertical distance between ballon and stone , y = 1896.22 - 27.1 * 17.1

y = 1432.8 m

hence , distance between stone and balloon = sqrt(335.2^2 + 1432.8^2)

distance between stone and balloon = 1471.5 m

the distance between stone and balloon is 1471.5 m

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