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!

To find the final velocity of the ball after 1 second, we can use the formula:

velocity (final) = velocity (initial) + acceleration x time

In this case, the initial velocity is 61 m/s, the acceleration is -9.8 m/s^2 (due to gravity), and the time is 1 second. Plugging these values into the formula:

velocity (final) = 61 m/s + (-9.8 m/s^2) x 1 s

Simplifying the equation:

velocity (final) = 61 m/s - 9.8 m/s^2

velocity (final) = 51.2 m/s

So, the ball is going at a speed of 51.2 m/s after being thrown for 1 second.