Question

Maximum height is the maximum vertical distance travelled by the projectile from the horizontal plane.$H=\left(\frac{u^2{\sin }^2\theta }{2g}\right)$ find distance.

Answer 1

Now in vertical direction $a=-g$
initial velocity =$u$
final velocity =$0$
using $v=u\tan$
$\sin \theta u=gt\therefore t=\frac{u\sin \theta }{g}$
now; distance travelled in this time in $y-dir$
$v^2-u^2=2as$
$u^2{\sin }^2\theta =2gs$
$\therefore s=\frac{u{\sin }^2\theta }{2g}=H_{max}$