A boy tosses a 0.5 kg ball straight up into the air.
How high up does the ball go if the ball leaves the boy's hand with a speed of 5 m/s?
Recall from the equations of motion

Where y is the maximum height achieved by the ball, v₀ is the initial velocity and t is the time.
Let us first find the time (t)

Re-arranging the equation for time (t)

Where v is the final velocity of the ball and is zero at the maximum height and g is the gravitational acceleration.

Now we can find the maximum height of the ball

Therefore, the maximum height of the ball 3.83 m