A ball is projected from point A with velocity 20 m/s prependicular to the inclined plane as shown in figure. What is the range of the ball on the inclined plane ?

Explanation

Direction down the inclined plane is taken as positive x -axis. The initial velocity of the ball is towards positive y-axis,.
` vecu = 20 hatj`
The acceleraton due to gravity can be also be resolved in two directions .
` veca = g sin theta hati - g cos theta hatj`
` 5 hat i - 5 sqrt3 hatj`
so,
` vecr = vecut + 1/2 veca t^(2)`
`= ( 20 t - 1/2 5sqrt3t^(2))hatj + 5/2 t^(2) hati`
when the ball hits inclined plane -again , its y - coordinate becomes zero ,so
`20 t - ( 5 sqrt3)/2 t^(2) = 0`
`or ( 5 sqrt3)/2 t = 20`
So ` t = 8/sqrt3 s`
Range is equal to x - coordinate of the paritcle
` R= 5/2 t^(2)`
` = 5/2 xx 64/3 = 160/3 m`
Range = `160/3` m