Question

On an inclined plane inclined at angle of 30 degree to horizontal , a ball is thrown upwards with velocity of 10 m/s , at angle of 60 degree to inclinedplane , its range on inclined plane is​

Answer 1

use formula of range of inclined plane, $R=\frac{2u^2sin(\alpha-\beta)cos\alpha}{gcos^2\beta}$

where α is angle of projection of particle with horizontal , β is inclination angle of plane with horizontal, u is initial speed of particle and g is acceleration due to gravity.

here, u = 10m/s, β = 30°

α = 60° to inclined plane. so, α = 60° + 30° = 90° with horizontal.

now, R = {(10)² × sin(90° - 30°).cos90°}/gcos²30°

we know, cos90° = 0

so, R = 0

hence range on inclined plane = 0 m

Answer 2

Answer:

0m

Explanation: