A ball player catches a ball 3.13 s after throwing it vertically upward. a) With what speed did he throw it?

b) What height did it reach?

a) v₀ = 15.3 m/s

b) h = 11.9 m

Explanation:

Ball Kinematics

We apply the free fall formulas:

vf = v₀+at Formula (1)

vf²=v₀²+2*a*y Formula (2)

y = v₀t + (1/2) * a*t² Formula (3)

y:displacement in meters (m)

v₀: initial speed in m/s

f: final speed in m/s

a: acceleration in m/s²

Data

t = 3.13 s

a=g = - 9.8 m/s² = acceleration due to gravity

Problem development

a) With what speed did he throw it

t = 3.13 s : total ball time going up and down The time the ball rises to its maximum height is half the total time At maximum height vf = 0

We apply the formula (1) for calculate initial speed (v₀)

vf = v₀+at

0 = v₀ + (-9.8) * (3.13/2)

v₀ = 15.3 m/s

b) What height did it reach?

We apply formula (2) or formula (3)

vf²=v₀²+2*a*h formula (2)

0 = v₀² + 2 * (-9.8) * h

19.6*h = v₀²

h=v₀²/19.6

h = (15.3) ² / (19.6)

h = 11.9m

y = v₀t + (1/2) * a*t² Formula (3)

h = (15.3) (3.13/2) + (1/2) (-9.8) (3.13/2) ² = 11.9m