A ball is thrown vertically straight upward with a speed of 48 ft/s. how long is it in the air

A ball is thrown vertically straight upward with a speed of 48 ft/s. how long is it in the air? Also how high will the ball be?

As the ball ascends, its velocity decreases, since it's against the pull of gravity, until it reaches 0 m/s. Then the ball with start descending, gaining velocity until it hits the ground, ignoring external forces.

Let's convert first ft to m.
48 ft/s * (30.48 cm / 1 ft) * (1 m / 100 cm) = 14.63 m/s

Vf = Vi + at
0 m/s = 14.63 m/s + (-9.8 m/s^2)t
-14.63 m/s = (-9.8 m/s^2)t
1.5 s = t

The ball will take 1.5 s to reach its maximum height. Therefore, we can also conclude that it will also take 1.5 s for the ball to return to the spot where it was thrown. So the total time of the ball in air is: 1.5 * 2 = 3 s

At the ball's maximum height, the final velocity is 0 m/s.
Vf^2 = Vi^2 + 2ad
(Vf^2 - Vi^2)/2a = d
([0 m/s]^2 - [14.63 m/s]^2)/2(-9.8 m/s^2) = d
10.9 m = d

Converting m to ft:
10.9 m * (100 cm / 1 m) * (1 ft / 30.48 cm) = 35.76 ft or 36 ft

Have a good day.

Please apply these formula:
For Time:
T=2VSin#/g
T=Time
V=Velocity
Sin#=Sin theta angle

For Height:

h=V^2.Sin^2#/2g

h= height
V=Velocity
Sin^2#= Sin square of theta.
g=gravity

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