A 124-kg balloon carrying a 22-kg basket is descending with a constant downward
ID: 1412550 • Letter: A
Question
A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 10.3 m/s A 1.0-kg stone is thrown from the basket with an initial velocity of 12.2 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 7.20 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 10.3 m/s .
A. At the instant the rock hits the ground, how far is it from the basket?
B. Just before the rock hits the ground, find its horizontal and vertical components as measured by an observer at rest in the basket.
C. Just before the rock hits the ground, find its horizontal and vertical components as measured by an observer at rest on the ground.
Explanation / Answer
Initial mass of ballon plus basket = 124 + 22 = 146 Kg
Final mass of ballon plus basket after 1 kg rick is thrown = 146 - 1 = 145 Kg
Initial downward velocity = final downward velocity = 10.3 m/s
Conservation of momentum in horizontal direction
v*1 + vbasket * 145 = 0
vbasket = v/145 m/s
vbasket + v = 12.2 m/s
=> 146v/145 = 12.2
=> v = 12.16 m/s
and horizontal vbasket = 0.084 m/s
A) time after throwing when it hits ground = 7.20 s
horizontal distance between rock and basket = 7.20 * 12.2 = 87.84 m
Vertical distance between rock and basket = 10.3 * 7.20 + 0.5*9.81*7.202 - 10.3*7.20 = 254.275 m
Total distance between rock and basket = 269m
B) With respect to someone in basket
Horizontal velocity of rock = 12.2 m/s
vertical velocity = 10.3 + 9.81*7.20 -10.3 = 70.632 m/s
C) With respect to somone stationary on ground
horizontal velocity = v = 12.16 m/s (discussed at the beginning of solution)
vertical velocity = 10.3+ 9.81*7.20 = 80.932 m/s
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