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.)(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Ball Thrown From Cliff

**Physics Forums | Science Articles, Homework Help, Discussion**