A rock is thrown from the edge of a building\'s roof with an initial velocity of

Question

A rock is thrown from the edge of a building's roof with an initial velocity of magnitude 8.0 m/s and a direction 53.1o above the horizontal. The rock hits the ground with a horizontal distance 48 m from where it was thrown. Neglect air resistance.

a) How long does it take the rock to travel from the roof to the ground?

b) What is the height of the building?

c) What is the maximum height avoe the roof reached by the rock during it's motion?

d) What is the speed of the rock just before it strikes the ground?

e) At what angle (relative to the horizontal) does the rock strike the ground?

Please show all work!

Explanation / Answer

a)

For horizontal motion,

D = ux*t

where D = 48 m

ux = horizontal speed = 8*cos(53 deg) = 4.81 m/s

So, t = 48/4.81 = 9.97 s <--------answer

b)

Using the equation of motion.

s = ut + 0.5*gt^2

For vertical motion,

u = uy = initial vertical speed = 8*sin(53.1 deg) = 6.4 m/s

g = -9.8 m/s2 <------- negative sign for downwards direction

So, s = 6.4*9.97 - 0.5*9.8*9.97^2 = -423.3 m

So, height of building = 423.3 m <--------answer

c)

Maximum height reached = 6.4^2/(2*9.8) = 2.1 m <-------answer

d)

vy = 6.4 - 9.8*9.97 = -91.3 m/s

vx = 4.81 m/s

So, speed of rock before striking ground = sqrt(91.3^2 + 4.81^2) = 91.4 m/s <------answer

e)

angle = atan(vy/vx) = atan(-91.3/4.81) = -87 deg <------answer

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