A 0.57 kg rock is projected from the edge of the top of a building with an initi

Question

A 0.57 kg rock is projected from the edge of the top of a building with an initial velocity of 7.97 m/s at an angle 51 degree above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 12.2 m from the base of the building. Assume: The ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s2 increased to the point that the centripetal acceleration was equal to the gravitational acceleration at the equator, what would be the tangential speed of a person standing a the equator? Answer in units of m/s.

Explanation / Answer

in horizontal,

t = 12.2 / (7.97 cos51 )

t = 2.43 s

in vertical,

y = v0 t + a t^2 /2

y = (7.97 sin51) (2.43) - (9.8)(2.43^2) / 2

y = - 14 m

so height of building = 14 m

-----------------------

a_c = v^2 / Re = g

v^2 = 9.8 x (6.37 x 10^6)

v = 7900 m/s

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