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...

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