A ball flies off the edge of a .68 m hightable. The ball hits the floor with a s

Question

   A ball flies off the edge of a .68 m hightable. The ball hits the floor with a speed of 6 m/s. How fast was the ball moving when it left the table? dy = v0 t + .5 (-9.8) (t)^2 to get t = .37s but I don't know how to get v0 with respect to x? Anyhelp is greatly appreciated.    A ball flies off the edge of a .68 m hightable. The ball hits the floor with a speed of 6 m/s. How fast was the ball moving when it left the table? dy = v0 t + .5 (-9.8) (t)^2 to get t = .37s but I don't know how to get v0 with respect to x? Anyhelp is greatly appreciated.

Explanation / Answer

We know from the relation v2 - u2 = 2gS where v and u are the initial and final velocities,g is theacceleration due to gravity and S is the height of the table. or u = (v2 - 2gS)1/2 where v = 6 m/s,g = 9.8 m/s2 and S = .68 m where v = 6 m/s,g = 9.8 m/s2 and S = .68 m

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