Question

A man stands on the roof of a building of height 16.7m and throws a rock with a velocity of magnitude 28.9m/s at an angle of 26.0❝ above the horizontal. You can ignore air resistance.

Part A

Calculate the maximum height above the roof reached by the rock.

Part B

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

Part C

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground

Answer 1

Answer 2

v = 28.9 m/s
a = 38 degrees
yo = 16.7 m
g = -9.81 m/s^2

vx = v * Cos(a) = 22.77 m/s
vy = v * Sin(a) = 17.79 m/s

tup = vy / g =1.815 s
h = yo + vy * tup + (1/2) * g * tup ^ 2 = 65.13 m

(1) height above roof:
y = 65.13-16.7 = 48.43 m

(3)
td = Sqrt(2 * h / g)= 1.846 s
T = Flight time = tup + td = 3.66 s
Range = vx * T = 83.36 m

(2)
vxf = 22.7 m/s
vyf = g * td = 18.09 m/s
V = Sqrt(vxf ^ 2 + vyf ^ 2) = 29.0265 m/s