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A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0 kg stone is thrown from the basket with an initial velocity of 15.0 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 6.00 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 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: i. At rest in the basket; ii. at rest on the ground.
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Asked 12/10/2008 9:20:51 AM
Updated 277 days ago|9/5/2025 9:55:54 AM
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a. Height of balloon when the rock was thrown
Vertical motion of the stone relative to the ground:
y_stone(t) = h v_b t (1/2) g t²
Substitute values:
0 = h (20)(6) 0.5(9.8)(6²)
0 = h 120 176.4
h = 296.4 = 296 m
Answer: Height of balloon when rock is thrown: 296 m
b. Height of balloon when rock hits the ground
y_balloon = h v_b t = 296.4 20(6) = 176.4 = 176 m
Answer: Height of balloon when stone hits ground: 176 m
c. Horizontal distance of stone from basket
x = v_s t = 15 × 6 = 90 m
Answer: Distance of stone from basket: 90 m
d. Horizontal and vertical velocity components just before hitting the ground
i. Relative to basket
Horizontal: v_x = 15 m/s
Vertical: v_y = g t = 9.8 × 6 = 58.8 59 m/s
Answer (relative to basket): v_x = 15 m/s, v_y = 59 m/s
ii. Relative to ground
Horizontal: v_x = 15 m/s
Vertical: v_y = v_balloon + v_stone relative to balloon = 20 + ( 58.8) = 38.8 = 39 m/s
Answer (relative to ground): v_x = 15 m/s, v_y = 39 m/s
Summary:
a) 296 m
b) 176 m
c) 90 m
d) Relative to basket: v_x = 15 m/s, v_y = 59 m/s; Relative to ground: v_x = 15 m/s, v_y = 39 m/s
Added 277 days ago|9/5/2025 9:55:54 AM
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