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 6.20 × 102 m above the ground and moving downward?

Answer 1

$h_m=\frac{v^2}{2g}=2580~m,$

$t_1=\frac vg=23~s,$

$t_2=\sqrt\frac{2(h_m-h)}g=20~s,$

$t=t_1+t_2=43~s.$