Determining Projection Angle With Little Information

[Ali0086]

Homework Statement

The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

2. The attempt at a solution

First I tried to find the speed for the max height.
v_f² - v_i² = -2gΔd
v_f² = v_i²- 2gΔd
0= v_i²- 2gΔd <--- Can someone explain why this must be set to zero?
∴ v_i=√(2gh)

After that I did the same for half the height

v_f² - v_i² = -2g(Δd/2)
v_f² = v_i² - g(Δd/2)
0= v_i² - g(Δd/2)
∴ v_i2=√(gh)

After reaching this point I'm not really sure how the given information can help give me an angle. I know somehow I'm supposed to break it down into horizontal components but I just can't see it. Any help would be appreciated.

 

[haruspex]

v_f² = v_i²- 2gΔd
0= v_i²- 2gΔd <--- Can someone explain why this must be set to zero?

That would follow if v_f=0, right? Will it be zero at maximum height? What will be zero at maximum height?

 

[phinds]

Forgetting the math for the moment, think about this. You are assuming, as one always correctly does in such basic problems, that air resistance is irrelevant and that the trajectory is ballistic. You also know that it's taking off and landing from the same height. What does all this tell you about the horizontal component of the "speed" at the max height vs at the launch? What does THAT tell you about the "speed" at the launch?

And, I assume that you understand the relationship between speed and velocity, yes?

 

[Ali0086]

That would follow if v_f=0, right? Will it be zero at maximum height? What will be zero at maximum height?

oh, let me guess, the final velocity will be zero right? The reason being there's that moment where it's absolutely still in the air.

 

[haruspex]

oh, let me guess, the final velocity will be zero right? The reason being there's that moment where it's absolutely still in the air.

If fired straight up, yes. Is this projectile fired straight up?

 

[Ali0086]

Forgetting the math for the moment, think about this. You are assuming, as one always correctly does in such basic problems, that air resistance is irrelevant and that the trajectory is ballistic. You also know that it's taking off and landing from the same height. What does all this tell you about the horizontal component of the "speed" at the max height vs at the launch? What does THAT tell you about the "speed" at the launch?

And, I assume that you understand the relationship between speed and velocity, yes?

Wouldn't that mean the horizontal component would remain constant, more specifically, the velocity of the horizontal motion wouldn't change.

 

[Ali0086]

If fired straight up, yes. Is this projectile fired straight up?

No it's an object in projectile motion. But what I'm getting at is that if the object is going on a projectile curve, for example a cannonball shot from a cannon. Wouldn't there be a moment in that parabolic motion where the velocity of the vertical component go from positive to negative? Doesn't that imply that the max height has been reached when that value of velocity is zero?

 

[phinds]

Wouldn't that mean the horizontal component would remain constant, more specifically, the velocity of the horizontal motion wouldn't change.

Yes and yes. Why would you expect otherwise. What would make it change?

 

[haruspex]

Wouldn't there be a moment in that parabolic motion where the velocity of the vertical component go from positive to negative? Doesn't that imply that the max height has been reached when that value of velocity is zero?

Yes, but only the vertical component. So what does v_i represent in your equation 0= vi²- 2gΔd?
The question refers to speeds. So what equations can you write down now?

 

[Ali0086]

the equation 1/2*(final velocity - initial velocity)
since 0= vi2- 2gΔd is the initial velocity

then isolate for the final velocity?

 

[haruspex]

the equation 1/2*(final velocity - initial velocity)

That's not an equation.

since 0= vi2- 2gΔd is the initial velocity

No, v_i is not "the initial velocity" in that equation in this context. Let's step back a moment. What is the 0 in that equation representing? Be exact.

 

[Ali0086]

well, I know vf2 - vi2 = -2gΔd represents the motion from height zero to the max height.
0= vi2- 2gΔd, so I believe the 0 here would be representing the fact that the vi and 2gΔd are equal right?

 

[haruspex]

well, I know vf2 - vi2 = -2gΔd represents the motion from height zero to the max height.
0= vi2- 2gΔd, so I believe the 0 here would be representing the fact that the vi and 2gΔd are equal right?

No, the equation represents that fact. The 0 is the value of a variable. What variable? Be as precise as you can.

 

[Ali0086]

alright, first off thanks for not just spoon feeding me answers. I haven't thought through physics like this in a while.
So being as literal as I can. 0= vi2- 2gΔd, looking at this equation, the first thing I can think of what that zero represents is the value of vf. So i assume it follows that vi2- 2gΔd point at which vf =0 .If that isn't right, thanks for trying. As of right now it's late where I live and I have classes tomorrow. I'll be back tomorrow to try and work through this. Thanks a lot though.

 

[haruspex]

alright, first off thanks for not just spoon feeding me answers. I haven't thought through physics like this in a while.

I'm glad you're taking my approach in a good spirit!
But I don't think we're going to get there at this rate. I've been trying to get you to echo back to me my statement that v_f in this equation represents the vertical component of the velocity at maximum height. On that understanding, what does v_i represent in that equation?

 

[Ali0086]

That would represent the horizontal component of the velocity at max height

 

[PeroK]

That would represent the horizontal component of the velocity at max height

I think a diagram of the motion would help a lot. This is a hard problem in my opinion and I would try to think out what's happening (without the equations of motion to begin with). So, I would draw a diagram showing:

The projectile as it leaves the ground; the projectile at its maximum height; the projectile when it has reached half its maximum height. I'd show the vertical and horizontal components of velocity at each of these points.

The next step is then to use the equations of motion to work out the velocity and/or speed at each point.

Also, as you have three points of interest here, it might be better to use $v_0, \ v_1\ \ v_2$ rather than $v_i, \ v_f$. In fact, keeping track of notation is a bit of a challenge in itself with this problem.

 

[haruspex]

That would represent the horizontal component of the velocity at max height

No, no! In the v_i, v_f notation, i stands for initial and f for final. As PeroK says, if there are multiple initial and final speeds to be considered (different directions, different phases of the action) it is a very good idea to use subscripts or a different letter to discriminate those. In this case, how about v₀, v₁, v₂ for the vertical speeds (v₂ being 0) and u (which does not change) for the horizontal speed?
In a given SUVAT equation, all the distances and speeds are in the same direction, so you apply the equations separately in (usually) the vertical and horizontal directions.
So please write down some equations relating to the vertical movement here.