Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago

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

Relevance
  • NCS
    Lv 7
    2 months ago
    Favorite 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.)

    If you find this helpful, please select Favorite Answer!

Still have questions? Get your answers by asking now.