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

  • Thread starter niki4d
  • Start date
13
0
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?)

---

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

Staff Emeritus
Science Advisor
8,056
72
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.
 
13
0
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

Mentor
44,810
1,074
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
Science Advisor
8,056
72
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!
 
Last edited:
13
0
Okay, Thank You! =)
 

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