Question

A 0.5 kg rock is projected from the edge of the top of a building with an initial velocity of 9.66 m/s at an angle of 41 degrees. The building is 10.1 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.

Answer 1

since you did not specify the direction very well (41 deg from horizontal or vertical? up or down from horizontal? Up or down from vertical?) I will have to be very general in the answer.
Vi = 9.66 sin or cos or -sin or -cos of 41

v = Vi - g t
at top v = 0
0 = Vi -9.81 t
so
t = Vi/9.81 at top

find h at top
h = 10.1 + Vi t - 4.9 t^2
solve for h, you know Vi and t

Now the rock drops from height h
0 = h + 0*Tfall - 4.9 Tfall^2
solve for Tfall, the time to fall from h

Now you know the rock has been in the air a time = t + Tfall

It had the same horizontal velocity the whole time

so
x = u ( t + Tfall)

where u = 9.66 cos or sin or -sin of 41