A ball, initially at the top of the inclined hill, is allowed to roll down. At the bottom its speed is 4 m/s. Next the ball is again rolled down the hill, but this time it does not start from rest. It has an initial speed of 3 m/s. How fast is it going when it gets to the bottom ?

Explanation

A ball start from the top of an inclined plane `u_(1)=0 m//s, v_(1)=4 m//s`
Let the acceleration 'a' and distance 's'.
`therefore " " v_(1)^(2)-u_(1)^(2)=2as " "` ………………….(1)
Now the ball is rolled down from the top of an inclined plane with a velocity 3 m/s.
`therefore u_(2)=3 m//s, v_(2)=?`
Here the acceleration of the inclined plane is 'a' and the distance is 's'.
`v_(2)^(2)-u_(2)^(2)=2as`
Solve (1) & (2)
`v_(1)^(2)-u_(1)^(2)=2as`
`(v_(2)^(2)-u_(2)^(2)=2as)/(v_(1)^(2)-u_(1)^(2)-v_(2)^(2)+u_(2)^(2)=0)`
`-v_(2)^(2)=-v_(1)^(2)+u_(1)^(2)-u_(2)^(2)`
Substitute the values `u_(1)=0` m/s, `u_(2)=3` m/s
`-v_(2)^(2)=-4^(2)+0-3^(2) rArr v_(2)^(2)=4^(2)+3^(2) rArr v_(2)^(2)=25 rArr v_(2)=5" "m//s`