Question

A 18.0-kg cannonball is fired from a cannon with muzzle speed of 1 050 m/s at an angle of 35.0° with the horizontal. A second ball is fired with the same initial speed at an angle of 90.0°. Let y = 0 at the cannon. (a) Use the isolated system model to find the maximum height reached by each ball.

Answer 1

To develop this problem we will apply the concepts related to energy conservation, in this way the change in kinetic energy must be equivalent to the change in potential energy. Finally we will obtain a speed dependent expression to find the maximum height that can be reached for the two given angles, given that the first cannon ball is fired with a speed of 1050m/s at angle of 35°.

Therefore the vertical velocity will be,

`v_y = (1005m/s)(sin35\°)`

Use the law of conservation of energy to find the height

`(1)/(2) mv_y^2 + 0 = mgh_(max) +0`

`v_y^2 = 2gh_(max)`

`h_(max) = (v_y^2)/(2g)`

Replacing with our values,

`h_(max) = ((1050*sin35\°)^2)/(2(9.89)`

`h_(max) = 18337m`

`h_(max) = 1.83*10^4m`

Therefore the maximum height reached by the first ball is
`1.83*10^4m`

The second cannon ball is fired at an angle of 90°. Hence, the maximum height reached by the second ball is,

`h_(max) = (v_y^2)/(2g)`

`h_(max) = ((1050sin90\°)^2)/(2(9.8))`

`h_(max) = 56250m`

`h_(max) = 5.6*10^4m`