A baseball is thrown from the rood of a 27.5m tall building with an initial velocity of magnitude 16.0m/s and directed at an angle of 37 degrees above the horizontal.

Question

A) Using energy methods and ignoring air resistance, calculate the speed of the ball just before it strikes the ground.

So, I did try out the problem using the formula 1/2mvf^2 +mghf = 1/2mvo^2 +mgho and I manipulated it until I got vf^2= sqrt(vo^2 +2g(ho-hf)

and the answer that I got for the final velocity before it hits the ground is 23.49 m/s. Is this correct?

Answer 1

I checked using the equations of motion. The height
h(t) = 27.5 + 9.629t - 4.9t^2
The ball hits after 2.237 seconds
The vertical velocity
v(t) = 9.629 - 9.8*2.237 = -12.294 m/s
The horizontal speed is 16cos37° = 12.778 m/s
So the final speed
s^2 = 12.294^2 + 12.778^2 = 314.424
s = 17.73

Answer 2

final energy = initial energy + energy from gravity

1/2 m Vf^2 = 1/2 m 16.0^2 + (m * 9.8 * 27.5)

Vf^2 = 256 + (2 * 9.8 * 27.5)