# 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...

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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### 1 Answer

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**