# Stone launched upward - Kinematics question

• EinsteinApple
In summary, Laura propels a stone straight up with her slingshot. The stone strikes a tree branch 4.0m above her. It strikes the branch with a velocity of 5.0m/s. It takes 1.6 seconds for the stone to hit the branch.
EinsteinApple

## Homework Statement

Laura propels a stone straight up with her slingshot. The stone strikes a tree branch 4.0m above her. It strikes the branch with a velocity of 5.0m/s. How long does this take?

a= -9.81 m/s2
v1 = 5.0m/s
d= 4.0 m[/B]

d=vit+1/2at2[/B]
d=v1+v2/2 x t
v2=v1+at

## The Attempt at a Solution

ATTEMPT 1: DOESN'T WORK
I believe I should use d=vit+1/2at2
therefore:
4m = (5.0)(t)+1/2(-9.81)(t)2
-4.91t2 + 5.0t -4 =0
However I can't do this because using quadratic I get a negative. Therefore it's not an option.

ATTEMPT 2: CAN I ASSUME v2=0?
d=v1+v2/2 x t
4=(5)+ (0)/2 x t
t = 1.6s
* But I guess I can't assume the V2=0

Attempt 3: CAN I ASSUME v2=0?
v2=v1+at
0=5 +(-9.81)t
t = 0.51 s.

I know the answer is 0.53s, however how do I know whether I should use attempt 2 or attempt 3?
I know to use attempt 3 because it's the right answer, but in my mind 1.6 s makes sense too.

If it hits the branch at ##5m/s## is that the initial velocity or the final velocity?

Think about the motion that you are studying. Where does it start and where does it end?

I believe it starts out of the sling at 5.0 m/s so that's v1. However it ends when it hits the branch? But it doesn't really, it will just bounce off and go in another direction.

EinsteinApple said:
I believe it starts out of the sling at 5.0 m/s so that's v1. However it ends when it hits the branch? But it doesn't really, it will just bounce off and go in another direction.

The "final" velocity is the velocity at the end of the motion you are interested in. In this case, when it hits the branch. You are not asked to study the motion of the stone after that.

The question says that the stone hits the branch at ##5m/s##. It does not say it leaves the slingshot at ##5m/s##.

Thank you! To clarify: the motion I am interested in is the rock flying from the slingshot. Therefore can I assume that v1=0? Or no because v1 is the rock out of the slingshot and therefore it would be a positive value?

EinsteinApple said:
Thank you! To clarify: the motion I am interested in is the rock flying from the slingshot. Therefore can I assume that v1=0? Or no because v1 is the rock out of the slingshot and therefore it would be a positive value?

If ##v_1 = 0##, how far do you think the stone would go? You are not given ##v_1##. It's an unknown quantity.

I understand that's in an unknown, I just assumed I could assume. Especially since my only two available equations have v1 in them.
d=v1+v2/2 x t
4=(unknown) + (5)/2 x t
t = 1.6s (I know this is incorrect)

v2=v1+at
5=unknown +(-9.81)t
t = -0.51 s. (I know this is correct but wrong sign)

EinsteinApple said:
I understand that's in an unknown, I just assumed I could assume. Especially since my only two available equations have v1 in them.
d=v1+v2/2 x t
4=(unknown) + (5)/2 x t
t = 1.6s (I know this is incorrect)

v2=v1+at
5=unknown +(-9.81)t
t = -0.51 s. (I know this is correct but wrong sign)

Both those equations have two unknowns: ##v_1## and ##t##. So, you may need to think of another equation!

So for each movement I should ask myself:
What is the segment of movement? In this case it's the rock out of a slingshot.
Therefore v1 is unknown because it's already launched out of the slingshot.
V2 = given
Because I am having difficulties knowing if v1=0 or v2=0 or neither

EinsteinApple said:
So for each movement I should ask myself:
What is the segment of movement? In this case it's the rock out of a slingshot.
Therefore v1 is unknown because it's already launched out of the slingshot.
V2 = given
Because I am having difficulties knowing if v1=0 or v2=0 or neither

Yes. In this case ##v_1## is unknown and ##v_2 = 5m/s##.

But, in this case, you are missing an important kinematic equation.

I understand! I should solve for v1, then use a secondary equation to find t.

EinsteinApple said:
I understand! I should solve for v1, then use a secondary equation to find t.

You can do it with the equations you have. But, there is another equation that would make things easier.

## 1. What is the initial velocity of the stone launched upward?

The initial velocity of the stone launched upward can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and t is the time. The initial velocity in this case would be equal to zero, since the stone starts from rest.

## 2. How high does the stone reach?

The height reached by the stone can be calculated using the equation h = ut + 1/2at^2, where h is the height, u is the initial velocity, a is the acceleration due to gravity, and t is the time. In this case, the final height reached by the stone would depend on the initial velocity and the time of launch.

## 3. What is the acceleration of the stone?

The acceleration of the stone is equal to the acceleration due to gravity, which is a constant value of 9.8 m/s^2. This means that the stone will accelerate downward at a rate of 9.8 m/s^2 for the entire duration of its flight.

## 4. How long does it take for the stone to reach the ground?

The time taken for the stone to reach the ground can be calculated using the equation t = √(2h/a), where t is the time, h is the height reached by the stone, and a is the acceleration due to gravity. However, this equation assumes that the stone is launched from the ground level and there is no air resistance. In reality, the time taken would be slightly longer due to the effects of air resistance.

## 5. How does the mass of the stone affect its trajectory?

The mass of the stone does not affect its trajectory, as long as the air resistance is negligible. This is because the acceleration due to gravity is independent of the mass of the object. However, a heavier stone would require more force to launch it upward, so a larger initial velocity would be needed for it to reach the same height as a lighter stone.

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