Question

A body is projected horizontally with a velocity of $10\text{m/s}$ from the top of an inclined plane of angle ${45}^{\circ }$. If it is to hit the foot of the plane and acceleration due to gravity is $10m/s^2$, the height of the plane (in $\text{m}$ ) is

Answer 1

Since angle of the plane is ${45}^{\circ }$ , vertical and horizontal distances covered by the body are equal.
i.e $h=x$.
We know distance covered by the body is given by, $x=u\sqrt{\frac{2h}{g}}$ where $u$ is the horizontal velocity.
$\Rightarrow u\sqrt{\frac{2h}{g}}=h\Rightarrow u^2\frac{2h}{g}=h^2$
$\Rightarrow h=\frac{2u^2}{g}=\frac{2\times {10}^2}{10}=20\text{m}$