Aball is rolled of a 2.14m table at 8m/s. find the time of flight for the ball and the horizontal range

0.661 s, 5.29 m

Explanation:

In the y direction:

Δy = 2.14 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.661 s

In the x direction:

v₀ = 8 m/s

a = 0 m/s²

t = 0.661 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²

Δx = 5.29 m

Round as needed.

answer: strength i guess

explanation:

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