A ball is thrown upward. The total time the ball remains in the air is 14 s. a) With what speed was the ball thrown upward? b) What is the maximum height reached by the ball?

a) Vi = 137.2 m/s

b) h = 960.4 m

Explanation:

a)

In order to find the initial speed we will use first equation of motion:

Vf = Vi + gt

where,

Vf = Final velocity = 0 m/s (since ball stops at highest point)

Vi = Initial Velocity = ?

g = - 9.8 m/s² (negative sign for upward moyion)

t = time interval = 14 s

Therefore,

0 m/s = Vi + (-9.8 m/s²)(14 s)

Vi = 137.2 m/s

b)

Now, we use second equation of motion to find height (h):

h = Vi t + (1/2)gt²

h = (137.2 m/s)(14 s) + (1/2)(-9.8 m/s²)(14 s)²

h = 1920.8 m - 960.4 m

h = 960.4 m