0.63kg rock is projected from the edge of the top of a building w/ initial velocity of 11.9m/s at an angle 59?

Question

...deg. above the horizontal. the building is 13.9m in height. accelertation of gravity 9.8m/s2. at what horizontal distance, from the base of the building will the rock strike the ground in meters?

Answer 1

- You don't need the mass of the rock.
Vertical:
v(zero y) = 11.9sin59 = 10.2m/s
s = s(zero) + v(zero) t + (1/2)gt^2
s = 13.9 + 10.2t - 4.9t^2
4.9t^2 - 10.2t - 13.9 = 0
t = (-B +/- √[B^2 - 4AC])/2A
t = (10.2 +/- √[(-10.2)^2 - 4(4.9)(13.9)])/9.8
t = (10.2 +/- 19.403)/9.8
The positive answer makes sense
t = 3.0207s
Horizontal:
D = rt
D = (11.9cos59)(3.0207)
D = 30.812m

Answer 2

My weight is exactly 63kg. If I am falling down at this rate of velocity & touch the base only calcium powder will remain!