Question

When a stone is thrown at angel 45 degree, with the horizontal it rises to a height 10m .What is the horizontal range

Answer 1

Given that
Angle of projectile = 45°
horizontal height = 10m
As we know that
$horizontal \: range \: = \frac{ {u}^{2} \times \sin2 \: theta }{2g} \\ \\ = \frac{ {u}^{2} \times sin \: 90}{2g} \\ \\ = \frac{ {u}^{2} }{2g}$
Also

MAX HEIGHT REACHED = u^2/2g

then
horizontal range = 10m