Question #15648

You throw a baseball directly upward at time=0 at an initial speed of 14.5 m/s. what is the maximum height the ball reaches above where it leaves your hand?

Expert's answer

At the highest point:

V = 0, V = V0 - g * t, so t = V0 / g

The maximum

height the ball reaches is:

h = V0 * t - g * t^2 / 2

h = V0 * V0 / g - g *

(V0 / g)^2 / 2 = V0^2 / (2 * g)

h = 14.5^2 / (2 * 9.81) = 10.7 m

V = 0, V = V0 - g * t, so t = V0 / g

The maximum

height the ball reaches is:

h = V0 * t - g * t^2 / 2

h = V0 * V0 / g - g *

(V0 / g)^2 / 2 = V0^2 / (2 * g)

h = 14.5^2 / (2 * 9.81) = 10.7 m

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