Question

A ball is thrown vertically upward, which is the positive direction. A little while later it returns to its point of release. The ball is in the air for a total time of 10 s. Note: Near the earth's surface, g is approximately 9.80 m /s2.

What is the algebraic expression for the initial velocity v0 of the ball? Express your answer in terms of the ball's displacement y, its acceleration a in the vertical direction, and the elapsed time t.

Answer 1

`v_0=(y)/(t)-(at)/(2)`

`v_0=49m/s`

Step-by-step explanation:

We can start from the known formula:

`y=y_f-y_0=v_0t+(at^2)/(2)`

so we can write, since we want the initial velocity in terms of the ball's displacement, its acceleration in the vertical direction, and the elapsed time:

`v_0=(y)/(t)-(at)/(2)`

and for our values we have:

`v_0=(0m)/(10s)-((-9.8m/s^2)(10s))/(2)=49m/s`