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 17.6 m/s
at an angle of 37 above the horizontal. The
rock strikes the ground a horizontal distance
of 74 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

Explanation / Answer

given velocity thrown=17.6 /_37 =>horizontal velocity Vh=17.6cos37=14.056 =>vertical velocity Vv=10.6 distance travelled horizontally=Vh*t=74 =>t=5.26 s final velocity vertical=gt+v g and Vv are opposite in direction Vvf=9.8*5.26-10.6=41 m/s magnitude=sqrt(Vvf^2+Vh^2)=43.34 m/s

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