Question

A STONE IS THROWN STRAIGHT UP INTO THE AIR AND HAS A HANG TIME OF 5 SECONDS FIND THE HEIGHT TO WHICH THE BALL REACHES ITS PEAK

Answer 1

Answer

Using newton second equation of motion

$v=u+at$

$0=u-gt$

u=gt

So maximum height

$H=\frac{u^2}{2g}\\=\frac{g^2t^2}{2g}\\=\frac{gt^2}{2}\\=\frac{10\times5\times5}{2}\\=125m$