Welcome to the forum. Thanks for posting your answers to the earlier parts of the problem. It's a great help in knowing what you do understand about the problem.

I started with a diagram and marked on the height of the cliff and the initial velocity upwards. This is 'up' but the cliff is down and so is the direction of gravity. So you need to decide at the start which way you will take as the positive direction and then use + or - correctly for the info given.

You've got the right formula to use but the signs aren't consistent. If you take 'up' as positive then u = 24.5, g = - 9.8, and s = - 117.6 (that's taking the cliff top as zero).

So the quadratic comes out as

If you take 'down' as positive then it is

As you can see these amount to the same thing so it doesn't matter which you use; but be consistent.

I tried this and got 8 seconds but not with your quadratic. I've just tried to solve your version with a quadratic solver and the square root part comes out negative so it cannot be evaluated. So did you make another sign error which 'cancelled out' the first?

For the last part the ball is to be 49m above the ground. But we are both measuring from the cliff top so you need to work out the distance down from the top to the 49 position ie 117.6 - 49. Then you can use that as your new 's' value to once again get a quadratic.

Hope that helps,

Bob

]]>I got -4.9t^2 + 24.5t - 117.6 = 0

19. Solve the equation you chose in question 18 to determine when the ball will hit the ground.

I got 8 seconds.

20. Using the same equation, determine when the ball is at a height of 49 meters.

Please help with number 20. This is my final question and I'm dying to get this done and over with. Thanks!

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