Question

A Particle is projected with a velocity 39.2 m/sec at an angle of 30o to an inclined plane (inclined at an angle of 45o to the horizontal). Find the range on the incline when it is projected upward.

• A. 114 m
• B. 112.5 m
• C. 113.7 m
• D. 166 m

Answer 1

The correct option is C 113.7 m

Time of flight on both cases will be same because the acceleration perpendicular to the plane is same. Therefore

$0=39.2\sin 30t-\left(1/2\right)g\cos 45t^{2}t=(2\times 39.2\sin 30)/\left(g\cos 45\right)=4\sqrt{2}sec$

Range upward$=39.2\cos 30t-\frac{1}{2}g\sin 30t^2=39.2\times \sqrt{3}/2\times 4\sqrt{2}-\left(1/2\right)\times 9.8\times \left(1/2\right)\times (4\sqrt{2}{)}^2=113.7m$