A 124- balloon carrying a 22- basket is descending with a constant downward velo

Question

A 124- balloon carrying a 22- basket is descending with a constant downward velocity of 11.4 A 1.0- stone is thrown from the basket with an initial velocity of 15.6 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 5.90 after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 11.4.

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?

Explanation / Answer

I assume standard units. When the stone is thrown it has 11.4 downward velocity and 15.6 horizontal velocity. It begins to accelerate downward. How far does a stone with an initial downward velocity of 11.4 fall in 5.90s d = vot + 1/2 gt^2 = 11.4*5.90 + (1/2)9.81(5.90^2) = 238 ANSWER Balloon was 238 high ANSWER b. Balloon is coming down at a constant 11.4 as the rock is accelerating to earth. ANSWER 238 - 11.4*5.90 = 170.7 ANSWER note that this is just the 1/2 gt^2 term from part a). The difference in the vertical position of the rock and balloon is just the contribution due to gravity. c. This whole time the rock is traveling at a constant horizontal speed of 15.6 Horizontal distance = 15.6*5.90 = 92 total distance is found by adding components using Pythagorian Theorem ANSWER how far from basket = SQRT(92^2 + 170.7^2) = 194

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