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:

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

Therefore:

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

Answer:

1.91m