Question

A body is projected from the bottom of a smooth inclined plane with a velocity of 20 $m{s}^{-1}$. If it is just sufficient to carry it to the top in 4 s, find the inclination and height of the plane.

Answer 1

Time taken to reach top$=4sec$

Let inclination of plane $\theta$ and length $h$ We know $v=u+at$

Final velocity, $v=0,t=4sec,a=-mg\sin \theta mg\sin \theta =\frac{2}{4}\Rightarrow mg\sin \theta =\mathrm{5...}\left(i\right)$

Also, $s=ut+\frac{1}{2}a{t}^{2}h=20\times 4-\frac{1}{2}mg\sin \theta (4{)}^{2}h=80-8\sin \theta mgmg\sin \theta =5[usingequation\left(i\right)]h=80-8\times 5h=40m$

By conservation of energy,

Gain in potential energy$=lossinK.Emg{h}_1=\frac{1}{2}m{v}^2{h}_1=h\sin \theta m\times 10\times 40\sin \theta =\frac{1}{2}m(20{)}^2\sin \theta =\frac{400}{2\times 40\times 10}\sin \theta =\frac{1}{2}\theta =30$