Posted by **Jennifer** on Thursday, September 14, 2006 at 3:44pm.

A ball is thrown from ground level up in the air at 61m/s. How fast is the ball going 1s after being thrown.

This problem wants you to use the formula

v = (-1/2)gt^2 + v_0

g=9.8m/s^2 and v_0 = 61m/s so

v = (-1/2)g*(1s)^2 + 61m/s at t = 1.

I wrote the formula wrong. It should be

v = -gt + v_0

I mixed the velocity formula and the distance formula.

You need the formula: velocity (finall) = velocity (initial) + at.

Here, the velocity initial is 61, and the t (time) is 1s. Remember, since the ball is decelerating (the acceleration is in the oppostie direction to which the ball is thrown), the a is -9.8m/s^2. Just plug 'em in!

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