A rock is pro jected from the edge of the top of a building with an initial velo

Question

A rock is pro jected from the edge of the top of
a building with an initial velocity of 18.1 m/s
at an angle of 35
?
above the horizontal. The
rock strikes the ground a horizontal distance
of 92 m from the base of the building.
The acceleration of gravity is 9.8 m/s
2
.
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 ?

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

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

Explanation / Answer

Vo=18.1 m/s==>=35o==>range=92 m==>ay=9.8==>vox=Vo*cos()14.83 m/s

voy=Vo*sin()10.4 m/s==>92=vx*t==>92/14.836.21 s==>vyfinal=voy+g*t71.2 m/s

vfinal=(71.22+14.832)72.72 m/s==>y=voy*t-1/2*g*t2-124.24 m or 124.24 m

vox=Vo*cos()14.83 m/s

ht=124.24 m

vyfinal=voy+g*t71.2 m/s

vfinal=(71.22+14.832)72.72 m/s

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