Ball Thrown From Cliff

  • Thread starter rdn98
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A man on the edge of a cliff H = 42 m high throws a ball directly upward. It returns past him 2 s later. (H actually is the height of the point of release of the ball above the base of the cliff. Neglect air resistance.) (Hint - Gravity causes a downward acceleration at the rate g = 9.81 m/s2.)

There are multiple parts, but I'm only stuck on this first part.
With what initial speed did the man throw the ball?

So at first, I thought it was zero, but computer told me it was wrong so its not. For this problem, I'm taking it to be upward is positive in the y direction, and downward past the man is negative.

I'm going to use the equation x-x0=v0(t)+.5(a)t^2
where x-x0=0
t=2sec
a=9.8m/sec^2

and I get v0=-9.8. The correct answer is 9.8. So what gives?
 

Answers and Replies

  • #2
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The a in your equation (the acceleration of gravity at the earth's surface) should be negative, if you want the direction down towards the center of the earth to be negative. This should fix the sign to be vo = 9.8 m/s.

Note: When doing problems draw a picture and label it with a frame, i.e. up/down (y-axis) and left/right (x-axis). Even when working with other areas of physics, it's critically important to place a frame of reference.

Frank
 
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