A marble is thrown horizontally with a speed of 10 m/s from the top of a buildin

Question

A marble is thrown horizontally with a speed of 10 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle 64° with the horizontal. From what height above the ground was the marble thrown?

Explanation / Answer

so assume the final velocity =v ==>vcos64=10 m/s (since there is no acceleration in the direction parallel to ground) ==>v=22.81 m/s so the vertical component of velocity =vsin64=20.50 m/s=vy(assume) so in direction perpendicular to ground we have u=0 ;v=20.50 m/s ;a=9.8 m/s2 ;assume s equation relating these is v^2-u^2=2*a*s ==> 20.50^2=2*9.8*s ==>s=21.44 m so height of building =21.44 m

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