A rock is projected from the edge of the top of a building with an initial veloc

Question

A rock is projected from the edge of the top of
a building with an initial velocity of 23.5 m/s
at an angle of 31 Degrees above the horizontal. The
rock strikes the ground a horizontal distance
of 88 m from the base of the building.
The acceleration of gravity is 9.8 m/s2 .
Assume: The ground is level and that the
side of the building is vertical. Neglect air
friction.
a) What is the horizontal component of the
rock’s velocity when it strikes the ground?
Answer in units of m/s
b) How long does the rock remain airborne?
Answer in units of s
c) How tall is the building?
Answer in units of m
d) What is the vertical component of the rock’s
velocity when it strikes the ground?
Answer in units of m/s
e) What is the magnitude of the rock’s velocity
when it strikes the ground?

Explanation / Answer

a.) Final horizontal velocity is the same as initial horizontal velocity (since it is not affected by the path of the rock) so: cos(angle) = Vx / V cos(31 degrees) = Vx / 23.5 Vx = 20.14 m/s b.) To find the time the rock is in the air, you can just use the horizontal distance along with the horizontal velocity you found in part a: d = V * t 88m = 20.14 m/s * t t = 3.97 seconds c.) To find the height of the building, you need the initial vertical velocity: sin(angle) = Vy / V sin(31 degrees) = Vy / 23.5 Vy = 12.10 m/s Then you can use that to find the net distance the rock would have traveled with the following equation: d = Vi*t + (1/2) * a * t^2 d = 12.10 * 3.97 + (1/2) * (-9.8) * 3.97^2 d = -29.19 m So the building is 29.19 meters tall. d.) To find the vertical component of the final velocity: Vf = Vi + a*t Vf = 12.10 + (-9.8)*3.97 Vf = -26.81 m/s So the rock's vertical velocity is 26.81 m/s straight down. e.) For magnitude of the final velocity: V^2 = Vx^2 + Vy^2 V^2 = 20.14^2 + 26.81^2 V = 33.53 m/s So magnitude of final velocity is 33.53 m/s.

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