A 0.75 kg rock is projected from the edge of the top of a building with an initial velocity of 11.9 m/s at an angle 59◦ above the horizontal. The building is 14.2 m in height. At what horizontal distance, x, from the base of the building will the rock strike the ground? Assume the ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s^2. Answer in units of m.
I do not care how massive it is.
horizontal speed the whole time = 11.9 cos 59 = 6.13 m/s
Now the real problem, vertical
Vi = initial speed up = 11.9 sin 59 = 10.20 m/s
Hi = initial height = 14.2
v = Vi - g t = 10.20 - 9.8 t
h = Hi + Vi t - 4.9 t^2
when is h = 0 (ground)
0 = 14.2 + 10.20 t - 4.9 t^2
solve quadratic for t, two solutions
t = -0.953 was before launch
t = 3.04 is useful
NOW for that whole 3.04 seconds the rock was going 6.13 m/s horizontal (remember?)
so
6.13 m/s * 3.04 s = 18.6 meters
I did that fast, check very carefully !