Apple and Peach?

You toss an apple horizontally at 9.2 m/s from a height of 2.7 m. Simultaneously, you drop a peach from the same height. How long does each take to reach the ground?

Apple and Peach?
Why would you do that to a perfectly good peach?





Seriously -- this is a classic problem.





The horizontal velocity is extraneous since they are both %26quot;falling%26quot; at 9.8 m/s^2 and will hit the ground at the same time, only the apple will be further away.





So surely you can figure the time it takes to fall 2.7 M.
Reply:Both will reach the ground at the same time. The fact that the apple has a horizontal velocity is not important. It is the vertical component of a velocity that counts in this situation.





The working formula is





S = VoT + (1/2(g)(T^2)





where





S = height at which both items started


Vo = vertical component of a velocity = 0


g = acceleration due to gravity (9.8 m/sec^2)


T = time for both items to reach the ground





Substituting values,





2.7 = 0 + (1/2)(9.8)(T^2)





Rearranging the above equation,





T^2 = (2)(2.7)/9.8 = 5.4/9.8 = 0.5510





T = 0.74 sec



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