A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 32 m/s at an angle of 50° with respect to the horizontal and the cannonball is 7.0 m above the ground at this time.
a) Assuming the cannonball makes it over the castle walls and lands back down on the ground, at what horizontal distance from its release point will it land?
b)What is the y-component of the cannonball's velocity just before it lands? The y-axis points up.
For this question, I did Vy=32sin50=24.5 (but this was also incorrect)
R= lvl^2/9 * (sin2theta)
The Attempt at a Solution
For part a, I used the equation for Range and got 102.9m but it was incorrect.
For part b, I did Vy=32sin50=24.5 (but this was also incorrect)