Object thrown at an angle: Initial velocity has to equal speed at t

[wicker558]

Homework Statement

Object is thrown at an angle of pi/4 (45°). What is the initial velocity if velocity at t = 2 seconds equals to initial velocity.

Relevant Equations

vi = vf
θ = 45°
t = 2 seconds

My solution is very sketchy, but we want the math right. However, I've came to 2 thoughts that helped me get to the solution, and those are:

• The only way that you can launch a ball at a velocity and have it be the same velocity two seconds later, is if it's already reached its maximum height and is now coming back down.
• Whatever velocity you began with is the "inclined" velocity, so you need to fix this.

The solution is 14 m/s. I would be glad if someone could post a way to solve this by deriving velocity as a function of time given the information we know.

 

[phinds]

The only way that you can launch a ball at a velocity and have it be the same velocity two seconds later, is if it's already reached its maximum height and is now coming back down.

Uh ... you might want to rethink that statement. Draw a trajectory diagram.

Whatever velocity you began with is the "inclined" velocity, so you need to fix this.

I have no idea what you mean here

 

[wicker558]

Uh ... you might want to rethink that statement. Draw a trajectory diagram.I have no idea what you mean here

I am a student, not a professor or physics expert. That's why I am seeking help. Constructive.

 

[etotheipi]

If this is projectile motion solely under gravity, the initial velocity won't be reached again but the initial speed might. There is no horizontal acceleration so we only need to impose the constraint that after 2 seconds, the vertical velocity is the negative of the initial vertical velocity in order for the magnitudes to be equal. Like @phinds mentioned, drawing a trajectory diagram makes this clear.

If the initial vertical component of velocity is $v\sin{\frac{\pi}{4}}$ and the final vertical component is $-v\sin{\frac{\pi}{4}}$ after a time interval of 2 seconds, can you think of a way of a way to put this information into a SUVAT equation?

 

[wicker558]

If this is projectile motion solely under gravity, the initial velocity won't be reached again but the initial speed might. There is no horizontal acceleration so we only need to impose the constraint that after 2 seconds, the vertical velocity is the negative of the initial vertical velocity in order for the magnitudes to be equal. Like @phinds mentioned, drawing a trajectory diagram makes this clear.

If the initial vertical component of velocity is $v\sin{\frac{\pi}{4}}$ and the final vertical component is $-v\sin{\frac{\pi}{4}}$ after a time interval of 2 seconds, can you think of a way of a way to put this information into a SUVAT equation?

Hey, sorry, meant the final speed. We call speed and velocity the same in my language.

 

[phinds]

Hey, sorry, meant the final speed. We call speed and velocity the same in my language.

If you are going to study physics, you must stop doing that. They are not the same and it will confuse you if you think they are.

 

[phinds]

I am a student, not a professor or physics expert.

Yes, I understand that. If you were, I would not have had to give you that hint. DRAW A DIAGRAM.

 

[wicker558]

If you are going to study physics, you must stop doing that. They are not the same and it will confuse you if you think they are.

I am completely aware that velocity is a vector and speed is a scalar. I am aware they are not the same. Just messed up the translation, the brighter ones get it. I profusely apologize.

 

[RPinPA]

I am completely aware that velocity is a vector and speed is a scalar. I am aware they are not the same. Just messed up the translation, the brighter ones get it. I profusely apologize.

Please don't be insulting, it will not help you get assistance here.

This is a teaching site. You have many knowledgeable and kind people here interested in helping students, but it is the expectation that the student will start the process by doing some of the work. If you are given a hint, you will find people more helpful if you follow that hint.

Out of curiosity, how in your language do you distinguish between the velocity vector and its magnitude, a scalar?

 

[archaic]

Since you know that $||\vec v(0)||=||\vec v(2)||$ from the HW Statement, you need to find $||\vec v(t)||=\sqrt{v_x^2(t)+v_y^2(t)}$.
A good place to start is writing down the accelerations $a_x(t)$ and $a_y(t)$ then integrating them. If you don't use calculus in your course, you should have a look at your SUVAT equations, there's one for velocities.

 

[jbriggs444]

A good place to start is writing down the accelerations $a_x(t)$ and $a_y(t)$ then integrating them.

No need to integrate. It is enough to note, as @etotheipi has mentioned, that the horizontal component of velocity is unchanging and that the way to get the speed back where it started is to negate the vertical component.

An obvious next question is: "When does that happen?"

A drawing could be helpful. Or not. It depends on whether one is more comfortable thinking visually or algebraicly.

 

[archaic]

No need to integrate. It is enough to note, as @etotheipi has mentioned, that the horizontal component of velocity is unchanging and that the way to get the speed back where it started is to negate the vertical component.

An obvious next question is: "When does that happen?"

A drawing could be helpful. Or not. It depends on whether one is more comfortable thinking visually or algebraicly.

You need to find $v_y$ somehow since you'd have the gravity term, i.e suvat or integration. I mention integration because I personally do not memorise the suvats nor do I look at them, I just start building from the zero.
That question is answered in the problem statement, $t=2$.

 

[jbriggs444]

I mention integration because I personally do not memorise the suvats

I agree in principle. But personally, I do not consider computing the integral of a zero function (and getting zero) or the integral of a constant function (and getting length times width) as "integration".

That question is answered in the problem statement, $t=2$.

Well noted. So the relevant question is "what would the initial vertical component of velocity need to be if the final vertical component of velocity is to be its negation at t=2".

That equation practically writes itself.

 

[phinds]

That equation practically writes itself.

Yes, but based on the OP's statements, it appears that he does not understand the situation at all which is why I repeatedly asked him to do a diagram.

 

[PeroK]

You need to find $v_y$ somehow since you'd have the gravity term, i.e suvat or integration. I mention integration because I personally do not memorise the suvats nor do I look at them, I just start building from the zero.
That question is answered in the problem statement, $t=2$.

You can integrate if you like, but I'd advise others not to follow your example.

 

[archaic]

You can integrate if you like, but I'd advise others not to follow your example.

why not? you'd just need to find the constant.

 

[PeroK]

why not? you'd just need to find the constant.

This is introductory homework and may be undertaken by those who have not yet learned the integral calculus.