Aball is thrown vertically upward with an initial speed of 18 m/s. then, 0.73 s later, a stone is thrown straight up (from the same initial height as the ball) with an initial speed of 30 m/s. how far above the release point will the ball and stone pass each other? the acceleration of gravity is 9.8 m/s 2 . answer in units of m.

Solution:

 the boll,  

y = Vo t + ½ g t²  

y = 18 t + ½ (-9.8) t²  

and the second one,  

y = Vo t + ½ g t²  

y = 30(t – 0.735) + ½ (-9.8)(t - 1.0.735)²  

the stone and the ball will pass each other if,  

y = y  

18 t + ½ (-9.8) t² = 30(t – 0.73) + ½ (-9.8)(t - .0.73)²  

t = 1.8196 sec  

y = 18 t + ½ (-9.8) t²  

y = 18(1.8196) + ½ (-9.8)(1.8196)²  

y = 16.5292 m above initial point

Thus, this the required solution.

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