# 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?

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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?

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#### cristo

Staff Emeritus
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
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution
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?

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

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

Mentor
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

Staff Emeritus
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?

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

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!

Last edited:

#### niki4d

Okay, Thank You! =)