Question

A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How long is required to reach this height?

For part A) I did: y=11m, g=9.81m/s^2, Vy0= 18 m/s, y0=0, (V^2)y=?

V^2 y= V^2 y0- 2g (y-y0)
--> V^2 y=18^2m/s-2(9.8)(11m-0m)
--> Vy=Square root ( 324m/s-215.6)= 10.4m/s

Part B) I started out with:
y=y0 + Vy0t - (1/2) gt^2
--> 11= 18t m/s-(1/2) 9.8t^2
--> -11 + 18t- 9.8t^2

Do I apply the quadratic formula? I do not know how to answer this, please help! This is due at midnight :(

Answer 1

Looks like it's after midnight already, but yes, you would need the quadratic formula for that last equation.

Answer 2

I still have about 53 minutes and I did that. But my answer is not coming out right according to the program. I know that one answer has to be positive and the other one negative since the stone passes the max height twice. But I plugged in the numbers and they both came out positive

Answer 3

I got: t=2.3205,-0.4837 this time. But the website said that I am either adding or missing a numerical constant. And It wants me to answer to three sig figs and I did. But it is not saying my numbers are wrong. Just the sig figs. So how would I put them down? I put 2.32, -0.482. Is this wrong?

Answer 4

Did you take half of 9.8?

Answer 5

I would think that both solutions should be positive: the first being when it reaches 11m on the way up and then the second when it gets there on the way down. It's only asking for the first time, I think.

Answer 6

The equation is\[-\frac{9.8m/s^2}{2}t^2+(18m/s)t-11m=0m\]
QF yields {0.7743s, 2.899s} as solutions.
From the wording of the question, I think it only wants the first time, 0.774s.

Answer 7

You are right! Thank you so much for clarifying! I re-did the equation for clarification and the answers ended up being right. Thank you so much for your help!