A small rock is thrown from the edge of the roof of a building with an initial v
ID: 1272853 • Letter: A
Question
A small rock is thrown from the edge of the roof of a building with an initial velocity that has magnitude of v=20 m/s and a direction of 53.1 degrees above the horizontal. The rock hits the ground 4 seconds after it was thrown. Neglect air resistance.
A) What is the horizontal distance from the point where the rock was thrown to where it strikes the ground?
B) What is the height of the building?
C) What is the maximum height about the ground reached by the rock during its motion?
D) What is the speed of the rock just before it strikes the ground?
Explanation / Answer
A) x = Vo*cos(theta)*t
= 20*cos(53.1)*4
= 40.03 m
B) h = vosin(53.1)*t - 0.5*g*t^2
h = 20*sin(53.1)*4 - 0.5*9.8*4^2
= 14.43 m
C) Hmax = h + voy^2/(2*g)
= 14.43 + (20*sin(53.1))^2/(2*9.8)
= 27.48 m
D) vy = voy - g*t
= 20*sin(53.1) - 9.8*4
= -23.2
vx = vo*cos(53.1) = 20*cos(53.1) = 12 m/s
V = sqrt(vx^2 + voy^2)
= sqrt(23.2^2 + 12^2)
= 26.12 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.