Physics question? I need help please!!

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Physics question? I need help please!!



A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 27.3 m/s A 1.0-kg stone is thrown from the basket with an initial velocity of 16.8 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 7.60 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 27.3 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.



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Since the stone is thrown horizontally, its initial vertical velocity is the same as the basket. Let’s use the following equation to determine the vertical distance it moves in 7.60 seconds. 

d = vi * t + ½ * g * t^2 
d = 27.3 * 7.60 + ½ * 9.8 * 7.60^2 = 490.504 meters 

B- How high is the balloon when the rock hits the ground? 
Let’s use the following equation to determine the vertical distance the balloon moves. 

d = v * t = 27.3 * 7.60 = 207.48 meter 
To determine the height of the balloon, subtract this distance from the maximum height 

d = 490.504 – 207.48 = 283.024 meters