Three balls are launched from the same point high on a cliff - they all end up l

Question

Three balls are launched from the same point high on a cliff - they all end up landing at different places on the flat ground below. The launch point is 28.0 m above the ground, and all the balls are launched with the same speed of 22.0 m/s.

Ball A is launched with an initial velocity that is entirely horizontal.
Ball B's initial velocity is directed at an angle of 37.0 degrees above the horizontal.
Ball C's initial velocity is directed at an angle of 26.0 degrees below the horizontal.
As usual, we will neglect air resistance. Use g = 9.80 m/s2.

Calculate the speed of object A, just before impact with the ground

Explanation / Answer

.5*g*t^2= 28 or, t=2.39s vertical velocity of A= g*2.39=23.42m/s horizontal velocity= 22m/s speed of A before impact= sqrt(23.42^2+22^2)=32.13m/s

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