Posted by Julia on .
A rock is projected from the edge of the top of a building with an initial velocity of 19.6 m/s at an angle of 31 degrees above the horizontal. The rock strikes the ground a horizontal distance of 77 m from the base of the building. The acceleration of gravity is 9.8 m/s^2.(squared) Assume: The ground is level and that the side of the building is vertical. Neglect air friction. What is the vertical component of the rock's velocity when it strikes the ground? Answer in units of m/s
Vo = 19.6m/s @ 31deg.
Xo = hor. = 19.6cos31 = 16.80m/s.
Yo = ver. = 19.6sin31 = 10.1m/s.
Dh = Xo*T = Hor. dist.
T = Dh / Xo = 77 / 16.8 = 4.58s = Time
t(up) = (Yf - Yo) / g,
t(up) = (0 - 10.1) / -9.8 = 1.03s.
t(dn) = T - t(up) = 4.58 - 1.06=3.55s.
Yf = Yo + gt,
Yf = 0 + 9.8*3.55 = 34.8m/s. = Ver.
component of final velocity.