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# A balloon, rising vertically with a velocity of 16 ft/sec, releases a sandbag at the instant when the balloon is 48 ft above the ground.?

Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)

A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 48 feet above the ground.

(a) How many seconds after its release will the bag strike the ground? (Round your answer to two decimal places.)

t = ? sec

(b) At what velocity will it strike the ground? (Round your answer to three decimal places.)

v = ? ft/sec

### 1 Answer

- NCSLv 72 months agoFavorite Answer
(a) solve y = y₀ + v₀t + ½at²

0 = 48 + 16t - 16t²

has roots at t = -1.3 s ← ignore

and t = 2.30 s ← solution

(b) v = v₀ + at = 16ft/s - 32ft/s²*2.3s = -57.689 ft/s

OR v = √(u² + 2gh) = √(16² + 2*32*48) ft/s = -57.689 ft/s

(I took the negative root to indicate down.)

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