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

In summary, the conversation discusses how to solve a problem involving a stone being thrown straight up with a given initial velocity and finding its velocity and time at a specific altitude. The conversation also mentions the use of kinematic equations and provides a method for solving the problem. It is determined that there are two possible answers as the stone passes the given altitude both on the way up and down. Finally, the correct procedure for finding the final velocity is confirmed.
  • #1
niki4d
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?
 
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  • #2
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.
 
  • #3
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..
 
  • #4
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.
 
  • #5
niki4d said:
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:
  • #6
Okay, Thank You! =)
 

1. How do you calculate the time it takes for a stone to reach a certain height when thrown straight up?

In order to calculate the time it takes for a stone to reach a certain height when thrown straight up, you can use the equation t = √(2h/g), where t is the time, h is the height, and g is the acceleration due to gravity (9.8 m/s²).

2. Does air resistance affect the time it takes for a stone to reach a certain height when thrown straight up?

Yes, air resistance does affect the time it takes for a stone to reach a certain height when thrown straight up. The more air resistance there is, the longer it will take for the stone to reach the desired height.

3. Can you use the same equation to calculate the time for a stone thrown straight up regardless of the initial velocity?

Yes, the equation t = √(2h/g) can be used to calculate the time for a stone thrown straight up regardless of the initial velocity. This is because the equation takes into account both the initial velocity and the acceleration due to gravity.

4. Is the time it takes for a stone to reach a certain height when thrown straight up affected by the mass of the stone?

No, the time it takes for a stone to reach a certain height when thrown straight up is not affected by the mass of the stone. This is because the acceleration due to gravity is constant and does not depend on the mass of the object.

5. Can this equation be used to calculate the time for any object thrown straight up, not just a stone?

Yes, this equation can be used to calculate the time for any object thrown straight up, as long as the object experiences a constant acceleration due to gravity and there is no other external force acting on it.

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