# A ball is thrown verticallyup at 10m/s.returns to ground with a velocity of 9m/s. What is the maximum height attained by the ball?

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For the motion of the ball we can apply equations of motion.

Initial velocity (U) = 10m/s

Final velocity at top (V) = 0m/s

Acceleration due to gravity (g) = 10m/s

Since the ball stops at maximum height and comes down we can say at the moment of maximum height the velocity of the ball is 0.

`uarr V^2 = U^2+2xxaxxS`

`0 = 10^2+2xx(-10)xxS`

`S = 5m`

*The ball reaches a maximum height of 5m from the level that it was thrown upward.*

*Note:*

*Since the ball gets a 5m max height we cannot say the height is from normal ground level.If it was thrown from normal ground level it should comes to ground at the same velocity as it was thrown. Here the velocity is lower than the initial when it comes to ground.So that means the ball has thrown from a level that is below the ground level.*

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