A stone thrown straight up how much time before it reaches a certain height?

[niki4d]

Here is my question:

A stone is thrown straight up with a speed of 15.0 m/s. How fast will it
be moving when its altitude is 8.0 m above the point from which it was
thrown? How much time elapses while the stone is reaching that height?
(Is there one answer or are there two answers? Why?)

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I know that the stone goes up and then down again, so there could be two answers, but how exactly would I solve this problem?

 

[cristo]

Hi and welcome to PF. In order for us to help you, you will need to show some working or your thoughts on the matter. Better still, as I presume this is a homework question, use the homework template
I'll give you a hint. You should use the "kinematic equations." Try googling, and giving it a go; post your attempts and we will be able to help further.

 

[niki4d]

Oh, okay, thank you.

Question:
A stone is thrown straight up with a speed of 15.0 m/s. How fast will it
be moving when its altitude is 8.0 m above the point from which it was
thrown? How much time elapses while the stone is reaching that height?
(Is there one answer or are there two answers? Why?)

Variables:
Vi = 15.0 m/s
d= 8.0m
a= -9.8 m/s/s
Vf = ?
t = ?

Equation I used:
d=vi(t) + 1/2 (a)(t)^2
Vf^2 = Vi^2 + 2ad

This is what I did;

8 = (15)t - 4.9 (t)^2
4.9t^2 - 15t + 8 = 0
Solving that, I got 2.37s, and 0.688s

For finding the final velocity (when at 8.0m);

(15)^2 + 2(-9.8)(8)
=8.26m/s

Is this the right procedure? Because I'm not sure if I'm doing this correctly..

 

[Doc Al]

Looks good to me. There are two answers since the stone passes the 8 m altitude both on the way up and on the way down.

 

[cristo]

Oh, okay, thank you.

Question:
A stone is thrown straight up with a speed of 15.0 m/s. How fast will it
be moving when its altitude is 8.0 m above the point from which it was
thrown? How much time elapses while the stone is reaching that height?
(Is there one answer or are there two answers? Why?)

Variables:
Vi = 15.0 m/s
d= 8.0m
a= -9.8 m/s/s
Vf = ?
t = ?

Equation I used:
d=vi(t) + 1/2 (a)(t)^2
Vf^2 = Vi^2 + 2ad

This is what I did;

8 = (15)t - 4.9 (t)^2
4.9t^2 - 15t + 8 = 0
Solving that, I got 2.37s, and 0.688s

Correct; the two answers appear since the stone moves past the point s=8m on the way up and on the way down.

For finding the final velocity (when at 8.0m);

(15)^2 + 2(-9.8)(8)
=8.26m/s

Is this the right procedure? Because I'm not sure if I'm doing this correctly..

Yes, that's correct.

Edit: Sorry, late!

 

[niki4d]

Okay, Thank You! =)