Acannon tilted up at a 30 angle fires a cannon ball at 80 m/s from atop a 10-m-high fortress wall. what is the ball's impact speed on the ground below?

To find the velocity of the bullet just before reaching the ground, we use the formula.
 Vy = Vosen (30 °) + gt (1).
 For that we need to know the time it takes for the bullet to reach the ground from the moment it is fired.
 For that we use the formula:
 Xy = Xo + Vosen (30 °) * t + 0.5gt ^ 2 (2)
 where:

 Vy = Component of the speed in the y direction (vertical)
 Vo = initial velocity of the bullet at the exit of the cannon
 g = acceleration of gravity
 Xy = vertical component of the position.
 Xo = initial position.
 t = time.

 To find the time it takes for the bullet to reach the ground, we make Xy = 0 and clear t.
 then it would be:
 0 = 10 + 80sen (30 °) * t +0.5 * (- 9.8) t ^ 2.
 The solution to that equation is:

 t = 8.4 seconds.

 We substitute that time in equation (1) and clear Vy.
 We have left:

 Vy = 80 * sin (30 °) + 9.8 * (8.4)
 Vy = 42.38 m / s

19 hertz bruh u trippin? ?