Question

g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find the ball's initial speed (in m/s).

Answer 1

The ball has height y and velocity v at time t according to

y = v₀ t - 1/2 g t ²

and

v = v₀ - g t

where v₀ is its initial speed and g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time t such that

-2.72 m = v₀ t - 1/2 g t ²

-7.94 m/s = v₀ - g t

Solve for t in the second equation:

t = (v₀ + 7.94 m/s)/g

Substitute this into the first equation and solve for v₀ :

-2.72 m = v₀ (v₀ + 7.94 m/s) /g - 1/2 g ((v₀ + 7.94 m/s)/g)²

-2.72 m = v₀²/g + (7.94 m/s) v₀/g - 1/2 (v₀ + 7.94 m/s)²/g

2 (-2.72 m) g = 2v₀² + 2 (7.94 m/s) v₀ - (v₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2v₀² + (15.9 m/s) v₀ - (v₀² + (15.9 m/s) v₀ + 63.0 m²/s²)

-53.3 m²/s² = v₀² - 63.0 m²/s²

v₀² = 9.73 m²/s²

v₀ = 3.12 m/s