A 124kg balloon carrying a 22kg basket is descending with a

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 12.5 m/s middot A 1.0-kg stone is thrown from the basket with an initial velocity of 18.2 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 14.9 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 12.5 m/s. How high was the balloon when the rock was thrown out? Express your answer with the appropriate units. h_0 = How high is the balloon when the rock hits the ground? Express your answer with the appropriate units. h_1 =

Solution

Let the required height be S.

We know from eqn of motion

S = ut + 1/2 at^2

In our case, S = ? ; u = vy = 12.5 m/s ; a = 9.8 m/s^2 ; t = 14.9 s

S = 12.5 x 14.9 + 1/2 x 9.8 x 14.9^2 = 1274.1 m

Hence, height = 1274.1 m

b)The balloon was at S = 1274.1 m

And its descended, h = 12.5 x 14.9 = 186.25

So the height when it hits the ground is:

H = 1274.1 - 186.25 = 1087.85 m

Hence, H = 1087.85 m