A balloon carrying a basket is descending at a constant velocity of 20.0 m/s. A person in the basket throws a stone with an initial velocity of 15.0 m/s horizontally perpendicular to the path of the descending balloon, and 4.00s later this person sees the rock strike the ground.
A. How high was the balloon when the rock was thrown out?
From h = Vot + gt^2, where h = height, Vo = initial downward velocity, g = gravity = 9.8m/s and t = time in seconds, h = 20(4) + 9.8(4^2)/2 = 158.4m.
B. How far horizontally does the rock travel before it hits the ground?
Dh = 4(15) = 60met.
C. At the instant the rock hits the ground, how far is it from the basket?
The balloon descends 80m in 4 seconds being 78.4m from the ground at the rock's impact. Therefore, the distance between the balloon and the rock is sqrt(78.4^2 + 60^2) = 98.72 m.
A balloon carrying a basket is descending at a constant velocity of 20.0 m/s. A person in the basket throws a stone with an initial velocity of 15.0 m/s horizontally perpendicular to the path of the descending balloon, and 4.00s later this person sees the rock strike the ground.
A. How high was the balloon when the rock was thrown out?
B. How far horizontally does the rock travel before it hits the ground?
C. At the instant the rock hits the ground, how far is it from the basket?
1 answer
2 answers