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

How tall is the building?
Answer in units of meters

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

Explanation / Answer

Since there are no forces in the horizontal direction, the velocity in the horizontal will be equal throughout

So by breaking the initial velocities into its initial x and y components you get
vy=vcos()=18sin40)

vy=11.57 m/s

vx=vsin()=18cos(40)

vx=13.79 m/s

Since the total horizontal distance is 94m and you know the horizontal velocity, so you can figure out the time traveled:

t=x/vx=94/13.79=t=6.82 seconds

Now we use the following equation to get the height

y=(1/2)gt2+vyt=(1/2)(-9.8)6.822+11.57(6.82)

y=-148.84

So the height is 148.84m tall

Finally, to get the vertical velocity when it hit the ground you use:

v=gt+vy

v=-9.8(6.28)+11.57

v=-55.23 m/s

Hope that helps

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