A ball is thrown down vertically with an initial speed of 20.5 m/s from a height of 58.8m.
What will be it's speed just before it strikes the ground? How long it will take for the ball to reach the ground?
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We have the initial speed: v0 = 20.5 m/s.
We also know the height from where the ball is thrown: h = 58.8m.
We'll write the equation of the motion:
v^2 = v0^2 + 2g*h
v = sqrt(v0^2 + 2g*h) (1)
We'll substitute the given data into the relation (1):
v = sqrt[(20.5)^2 + 2*9.8*58.8]
v = sqrt(420.25 + 1152.48)
v = sqrt(1572.73)
v = 39.7 m/s
The speed of the ball before striking the ground is v = 39.7 m/s.
To calculate how long it will take for the ball to reach the ground, we'll write the equation of motion:
v = v0 + g*t
We'll subtract v0 both sides:
gt = v - v0
We'll divide by g:
t = (v-v0)/g (2)
We'll substitute the velocities in (2):
t = (39.7 - 20.5)/9.8
t = 19.2/9.8
t = 1.96 s
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