# Suppose a ball is thrown straight in the air. If it takes the ball 4.8 s to return to its initial position, what was the initial velocity of the ballPlease help!

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Using straight-line motion, a ball thrown up in the air has the equation of motion given by:

`h=-1/2g t^2+v_0t+h_0` where `g=9.8` is acceleration due to gravity, `v_0` is the initial velocity and `h_0` is the initial height.

Using the height equation, when the ball comes back down, we have the height is the same as the initial height. Substituting the time into the height equation, we get:

`h_0=-1/2(9.8)(4.8)^2+v_0(4.8)+h_0` cancel `h_0` and simplify

`1/2(9.8)(4.8)^2=v_0(4.8)` divide by 4.8

`4.9(4.8)=v_0`

`v_0=23.52`

**The initial velocity of the ball is 23.52 m/s.**

The case is simply the same as dropping a ball from some height and what is the velocity when it hits the ground after travelling a time half of the time given in your problem that is 4.8/2 s, so t = 2.4 s.

so applying: v(t) = 0 + g t = 9.8 * 2.4 m/s = 23.52 m/s