Question

A cannon shell is fired straight up from the ground at an initial speed of 225 m/s. After how much time is the shell at a height of 102m above the ground and moving downward?

Answer 1

$y=y_0+v_0t+\frac{gt^2}{2}$

$102=0+225\cdot t-\frac{9.81\cdot t^2}{2}\to 4.905t^2-225t+102=0\to$

$t=45.4(s)$