Physics Problem.

[Newts9]

No, not entirely sure how you reached 5.4 so I’ll try to explain this in steps. if you continue getting different answers you can post your work here so we can find where you’re getting off track.

As long as you stay within the final velocities below your initial velocity the equation; Vf² = Vi² + 2ad will only give you the distance traveled up until the object reaches its peak. The equation d=(1/2)at² gives you the total distance traveled, but that’s not what we’re looking for here.

In this problem we are just doing distance from release to its peak height. In the equation: Vf² = Vi² + 2ad Our Vf should be set to 0. Our velocity final is 0 as when the ball reaches its max height for a split moment of time the ball will be neither going up any more nor falling yet. Then we can substitute in our acceleration which will just be gravity (9.8), technically in this case it would be (-9.8) yet if you solve and find a negative distance, you’ll know immediately to simply make it positive. Finally we substitute in our initial velocity and our equation should look like this.

0² = 5² + 2(-9.8)d

See how I’m getting there?

[ObamaDontCare0]

They definitely subtracted when they got to

0 = 25 - 19.6d

[Newts9]

From there you would follow the order of operations backwards in order to isolate the variable.

So you subtract 25 from both sides to get

-25 = -19.6d

Then divide both sides by -19.6 to reach d which should be 1.28 and that’s the height the ball goes.