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 18 m/s
at an angle of 40 above the horizontal. The
rock strikes the ground a horizontal distance
of 94 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.
What is the horizontal component of the
rock’s velocity when it strikes the ground?
Answer in units of m/s

How tall is the building?
Answer in units of m
016 (part 4 of 5) 10.0 points
What is the vertical component of the rock’s
velocity when it strikes the ground?
Answer in units of m/s
017 (part 5 of 5) 10.0 points
What is the magnitude of the rock’s velocity
when it strikes the ground?
Answer in units of m/s

How long does the rock remain airborne?
Answer in units of s

Explanation / Answer

convert the velocity into x y component.
velocity = 18cos40i + 18sin40j
= 13.789i + 11.57j

find the time taken for it to reach the ground :
94/13.789 = 6.82 seconds

find the time it takes to reach max. height:

v=0

t = ( Vf - Vi ) /a

t = (0-11.57)/-9.8

= 1.18 seconds

therefore, time taken for rock from max height to reach bottom:

6.82 - 1.18 = 5.64 seconds

velocity = 0 + 9.8*5.64

= 55.272 m/s

height of building = (Vf² - Vi²)/2a

= (55.2722 - 11.572)/ (2*9.8)

= 149.04 m

magnitude = (55.2722 + 13.7892) = 56.97 m/s

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What is the horizontal component of the
rock’s velocity when it strikes the ground?
Answer in units of m/s 13.789 m/s

How tall is the building?
Answer in units of m 149.04 m
016 (part 4 of 5) 10.0 points
What is the vertical component of the rock’s
velocity when it strikes the ground?
Answer in units of m/s 55.272 m/s
017 (part 5 of 5) 10.0 points
What is the magnitude of the rock’s velocity
when it strikes the ground?
Answer in units of m/s 56.97 m/s

How long does the rock remain airborne?
Answer in units of s 6.82 seconds

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