# A ball has to be thrown on a ledge 5 m above and 60 m away. What is the minimum speed with which it should be thrown.

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### 1 Answer

The ball has to be thrown on a ledge that is 5 m above the point from which it is being thrown and 60 m away. Let the speed at which the ball is thrown be v and the angle with the horizontal at which it is thrown be A.

The velocity of the ball can be divided into two components, one parallel to the horizontal and the other perpendicular to the horizontal. The first component is `v*cos A` and the other is `v*sin A` . When the ball is thrown there is an acceleration due to gravity equal to 9.8 m/s^2 in a vertically downwards direction. The minimum speed with which the ball can be thrown is one that ensures the highest point of the trajectory is 5 m and lies 60 m away from the point where the ball is thrown.

The vertical height of the ball after it has traveled 60 m is given by `60*tan A - (9.8*60^2)/(2*v^2*cos^2A)` .

`60*tan A - (9.8*60^2)/(2*v^2*cos^2A) = 5`

=> `60*tan A - 17640/(v^2*cos^2A) = 5`

=> `60*tan A - 5 = 17640/(v^2*cos^2A)`

=> `v^2 = 17640/(cos^2A*(60*tan A - 5))`

To minimize v^2 , `cos^2A*(60*tan A - 5)` has to be maximized.

Using a graphing calculator the maximum value is determined to be 27.603986447. This gives the minimum value of v as approximately 25.27 m/s.

**The minimum speed with which the ball can be thrown is approximately 25.67 m/s.**

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