Determine the minimum initial velocity

[derek181]

Homework Statement

Determine the minimum initial velocity V0 and the corresponding angle theta at which the ball must be kicked in order for it to just cross over the 3 m fence.

Given:
Horizontal distance=6m
vertical distance=3m

Homework Equations

equation 1: V_y²=(V₀)²+2a(y-y₀)

equation 2: V_y=(V₀)_y+at

equation 3: X=X₀+(V₀)_xt

The Attempt at a Solution

Using equation 1
0=V_0y+(2)(-9.81)(3-0)
V_oy=7.67m/s
using equation 2
0=V_0y+(-9.81)t
t=.782s
using equation 3
6=0+V_0x(0.782)
V_0x=7.67m/s
V₀=√(7.67²+7.67²)
=10.85m/s
θ=45°

 

[phinds]

You need to SHOW your work, not just describe it.

 

[derek181]

I see so many views. Does nobody really know how to solve this?

 

[phinds]

I see so many views. Does nobody really know how to solve this?

I'm confident there are thousands of people on this forum who know how to solve this, but that is not the way this forum works. SHOW YOUR WORK and we'll help you figure out where you are going wrong. We do not spoon-feed answers. Read the forum rules.

EDIT: OOPS ... I now see that you went back and DID show your work.

 

[derek181]

Care to tell me where I am going wrong Phinds?

 

[Dick]

Care to tell me where I am going wrong Phinds?

I can tell you where you started to go wrong. In equation 1, $V{_y}^2$ should be $V^2$ where V is the speed, not the y component of velocity. You can't split that kinetic energy equation into components. So when you got 7.67m/s, that's the initial speed, not the y component of velocity.

 

[derek181]

All I am doing here is applying equation 1 in the vertical direction which is valid. It even states in my textbook that for the vertical motion component V_y²=V₀²-2g(y-y₀. The opening section in the "motion of a projectile" section in the textbook states that the free flight motion of a projectile is often studied in terms of its rectangular components. The answer is 9.76m/s, I just have no idea how they got that answer.

 

[Dick]

All I am doing here is applying equation 1 in the vertical direction which is valid. It even states in my textbook that for the vertical motion component V_y²=V₀²-2g(y-y₀. The opening section in the "motion of a projectile" section in the textbook states that the free flight motion of a projectile is often studied in terms of its rectangular components. The answer is 9.76m/s, I just have no idea how they got that answer.

Do you mean you don't know how they got that equation? I do, it's conservation of energy, and given in terms of y components it is only valid if the motion is purely vertical, i.e. $V_x=0$. Not true here. Motion of a projectile is often split into components, but that equation doesn't split.

 

[haruspex]

Care to tell me where I am going wrong Phinds?

Apart from a missing power of 2 in "using equation1", it all looks fine. Why do you think it's wrong?

 

[haruspex]

Do you mean you don't know how they got that equation? I do, it's conservation of energy, and given in terms of y components it is only valid if the motion is purely vertical, i.e. $V_x=0$. Not true here. Motion of a projectile is often split into components, but that equation doesn't split.

It does become conservation of energy when you multiply by mass, but it is also a standard SUVAT
equation, valid for constant acceleration. It works just fine in each coordinate. So, in a way, it looks like work is conserved separately in each coordinate in such motion. That shouldn't be surprising since work should still be conserved for an observer moving with constant relative velocity.

 

[derek181]

I understand where the formula comes from. I derived it using calculus. Dick says you can't use this equation to analyze only the vertical motion and haruspex says you can? My Hibbler engineering mechanics textbook says you can analyze using components. Haruspex, the answer I've shown is incorrect.

 

[haruspex]

Haruspex, the answer I've shown is incorrect.

Then are you sure you have stated it correctly. The 6m is the distance to the fence, right?

 

[derek181]

Copied the question word for word. 6m horizontal distance to the fence. This one is a tough one.

 

[haruspex]

Copied the question word for word. 6m horizontal distance to the fence. This one is a tough one.

Then I see no way your answer can be wrong.

 

[derek181]

That's what I think as well but the answer is 9.76m/s with an angle of 58.3degrees. There must be someone on this board who knows how to solve it.

 

[derek181]

I figured it out. There is a whole lot more to this problem than just those simple kinematic equations. We use x=x0+v0xt and y=y0+v0t+.5gt. After subbing in all values we will have two equations and two unknowns. So we solve for V0. We will then have a term in the denominator of the square root that is sin(2theta)-Cos^2(theta) and the trick is to find when theta is a maximum. So we do a bit of calculus and then find the angle which gives Vo as a minimum. Thanks anyway. This one was a tough question I guess.

 

[haruspex]

That's what I think as well but the answer is 9.76m/s with an angle of 58.3degrees. There must be someone on this board who knows how to solve it.

Doh! You and I made the same mistake - we just assumed that the least speed was the one that meant it was traveling horizontally as it cleared the fence. That's a bad assumption.
Start again, using the general equations for the trajectory, specifying that it passes through the point (6, 3), and using calculus to find the least speed.

Edit: I see you got there about the same time.

 

[Dick]

It does become conservation of energy when you multiply by mass, but it is also a standard SUVAT
equation, valid for constant acceleration. It works just fine in each coordinate. So, in a way, it looks like work is conserved separately in each coordinate in such motion. That shouldn't be surprising since work should still be conserved for an observer moving with constant relative velocity.

Yeah, sorry. I was so tired last night, I honestly don't remember what I was thinking. Bad time to answer questions.