10) A bowling ball released from rest at the top of a 200. m building falls to the ground under the influ-

ence of gravity. Calculate:

a) its speed at 5.00 s

b) how far it has gone when its speed is 32.0"

Question

Taliyah Murillo

Physics

21 December, 07:38

10) A bowling ball released from rest at the top of a 200. m building falls to the ground under the influ-

ence of gravity. Calculate:

a) its speed at 5.00 s

b) how far it has gone when its speed is 32.0"

+4

Answers (1)

Know the Answer?

Marley · Answer 1

Given:

v₀ = 0 m/s

a = 9.8 m/s²

a) Find v when t = 5.00 s.

v = at + v₀

v = (9.8 m/s²) (5.00 s) + 0 m/s

v = 49.0 m/s

b) Find Δx when v = 32.0 m/s.

v² = v₀² + 2aΔx

(32.0 m/s) ² = (0 m/s) ² + 2 (9.8 m/s²) Δx

Δx = 52.2 m

10) A bowling ball released from rest at the top of a 200. m building falls to the ground under the influ-ence of gravity. Calculate:a) its speed at 5.00 sb) how far it has gone when its speed is 32.0"

10) A bowling ball released from rest at the top of a 200. m building falls to the ground under the influ-

ence of gravity. Calculate:

a) its speed at 5.00 s

b) how far it has gone when its speed is 32.0"