A cannon fires a shot at an angle of 60 degrees with a velocity of 113m/s. If the cannon is 49 meters above sea level what is the highest hight the cannonball will reach before it hits the water?

I started this problem and found that the time it would take is 20.5 seconds and that the velocity in the y direction is 97.9m/s but I'm stuck on how to find the maximum height (I've only found it when the change in why was zero)

Hallo !

Just use this link to learn how to solve this problem. (http://www.makingthemodernworld.org.uk/learning_modules/maths/04.TU.02/?section=12)

Succes !

:)

Break the velocity of the canon ball into vertical and horizontal components. You are really only interested in the vertical component, since the question asks nothing about how far the canon ball travels. The initial vertical component of velocity is:


v_i = 113 \frac m s \cdot sin(60)


Now use this initial velocity with the equation:

{v_f}^2 - {v_i}^2 = 2ad


If you set v_f equal to zero, and use a = -g and v_i = 113 m/s * sin (60), you can calculate d . This is the vertical distance traveled for the canon ball to reach a vertical velocity of 0, which is when it's at the top of its projectory. Don't forget that all this is calculated from a starting point of 49m, so you will need to add that in.