Question

A ball is thrown up into the air for a total of 1.25 s before it is caught at its original position . How high did the ball go

Answer 1

`v^2=v^2_o+2g(y-y_o)`

so:

The initial speed is given by:

`\begin{gathered} v=v_o+at \\ _{\text{ }}where\colon \\ v=-v_o \\ a=g=-9.8 \\ t=1.25 \\ so\colon \\ -v_o=v_o+(-9.8)\cdot1.25 \\ -2v_o=-12.25 \\ v_o=(-12.25)/(-2) \\ v_o=6.125 \end{gathered}`

Therefore:

`\begin{gathered} y_o=0 \\ v_{}=0 \\ v_o=6.125 \\ so\colon \\ y=(-v^2_o)/(2g) \\ y=(-6.125^2)/(2(-9.8)) \\ y=1.91m \end{gathered}`

Answer:

1.91m