Question

A particle is projected at angle ${37}^o$ with the incline plane in upward direction with speed $10m/s$. The angle of incline plane is given ${53}^o$. Then the maximum distance from the incline plane attained by the particle will be

• A. $3m$
• B. $4m$
• C. $5m$
• D. $zero$

Answer 1

The correct option is A $3m$

$x=\frac{u^2}{2g}=\frac{{10}^2}{2\times 10}=5m$

As $h$ is the perpendicular distance from the incline plane

$\sin {37}^o=\frac{3}{5}$

or, $\frac{h}{x}=\frac{3}{5}$

$\therefore h=\frac{3}{5}\times x=\frac{3}{5}\times 5=3m$

Here range on inclined plane will be zero, as particle goes vertically upand comes back.