Originally Posted by

**stapel** If you mean that h(0.5) = 3.75, then I agree with your answer.

No i mean 0.5 seconds. "Find the height of the ball after 0.5s" - Don't you make this 3.75 then?

Mr F says: Stapel has agreed with you! She worded it like this because the way you expressed your answer was poor. Originally Posted by

**stapel** Cant i just find the answer of the graph? In which case when h=0, the two times are 2secs and ?

Mr F says: You've been told how to do it .... Originally Posted by

**stapel** Since the ball is assumed to hit the ground and stop (rather than continuing downward through the Earth), what is the last t-value for which the h-values make physical sense? (You'll use your answer to part (2) for this.)

How do i know where the ground is though?

Mr F says: It should be clear that the ground corresponds to h = 0. Originally Posted by

**stapel** I'm not sure what they mean by their first "method", unless they're wanting you to

**complete the square** to find the vertex...?

Using the formula is often easier, and you might want to do "method 2" first, so you can be sure of your answer to "method 1". You might want to check your work on this, by the way: -(10) does not equal "10".

So are my co-ordinates wrong? - If so what do you make them, and could you please explain your method?

Mr F says: Did you check your work like was was suggested? Originally Posted by

**stapel** Yes, the velocity at the peak of its path is zero; it has stopped going up, and is about to come back down.

Woo!

Originally Posted by

**stapel** How did you arrive at 10 - 10(1/2) equalling 1? Or 10 - 10(3/2) equalling -1?

Ok, i have no idea what i did, so we have $\displaystyle v = 10 -10t$ - "Find the velocity of the ball @ t = 0.5s and t = 1.5s

Mr F says: Stapel is saying, kindly, that you have botched the basic arithmetic. Go back and do it right. Originally Posted by

**stapel** A positive velocity indicates forward (in this case, upward) motion; a negative velocity indicates backward (in this case, downward) motion.

Originally Posted by

**stapel** The speed is the velocity, but without regard to direction. You'd take the absolute value, not the derivative.

Is there a formula i could show this working?

Mr F says: Do what stapel said to do. Originally Posted by

**stapel** Just follow the instructions: Plug "1.8" in for "h", which means "the times t when the height h is 1.8 meters", and rearrange (by putting everything together on one side of the "equals" sign and dividing through by the appropriate value) to get the equation into the form they showed.

Ok so it looks like $\displaystyle t^2 - 2t + 0.36 = 1.8$ where do i go from here?

Mr F says: No! Where did you get this from? You certainly didn't follow the instruction very carefully. Substitute h = 1.8 into h = -5t^2 + 10t: -5t^2 + 10t = 1.8 => 5t^2 - 10t + 1.8 = 0. Now divide both sides by 5.
Thanks for your help