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Projectile Motion above Ground

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Homework Statement



3) A projectile is launched with speed v₀ at an angle α₀ above the horizontal. The launch point is a height H above the ground. Ignore air resistance.

(a.) Find the horizontal velocity of the projectile.
(b.) Find the time for projectile to reach its maximum height.
(c.) Find the maximum height above the ground the projectile reaches.
(d.) Find the time it takes the projectile to fall to the ground from its maximum height.
(e.) Find the horizontal distance the projectile travels before striking the ground.

Homework Equations



I'm generally taught not to memorize any equations, but I'm capable of deriving five kinematic equations, and know enough trigonometry to make use of vector components

The Attempt at a Solution



I've actually answered parts (a.) through (c.), but I can't figure out the rest because I don't exactly know how I can take into account the extra height given by the launch height H.
 

Answers and Replies

  • #2
Bandersnatch
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The position equation you're using seems to be missing something.
If you get your equations through integration, remember to add a constant at the end of each step.

If you can't do integration, ask yourself: what do I need to add to my position equation to make it equal H at t=0?
 
  • #3
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Something is missing?

Well, in any case, I thought I should go ahead and post my answers to the previous questions to see if they could help lead me anywhere

(a.) v0cos[itex]\alpha[/itex]0 = v0x

I got this by using vector components and the given angle. The horizontal component is adjacent to the given angle, so I used the cosine equation

(b.) t = [itex]\frac{v_{0}sin\alpha_{0}}{g}[/itex], where g is the acceleration due to gravity.

This is obtained by using the kinematic equation: v = v0y + at, plugging in the vertical component for the initial velocity

(c.) [itex]\frac{(v_{0}sin\alpha_{0})^{2}}{2g}[/itex] + H = Maximum height above ground

I obtained this the same way I obtained (b.) only this time, I used a different kinematic equation: v2 = [itex]v^{2}_{0}[/itex] + 2a[itex]\Delta[/itex]r

The added H should take into account the missing height from the ground to the launch height.

So if the launch height was equal to ground level, I could just take the time it took to reach the maximum height since the motion should be symmetrical about the max height. Could I use that information somehow?
 
  • #4
SammyS
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Something is missing?

Well, in any case, I thought I should go ahead and post my answers to the previous questions to see if they could help lead me anywhere

(a.) v0cos[itex]\alpha[/itex]0 = v0x

I got this by using vector components and the given angle. The horizontal component is adjacent to the given angle, so I used the cosine equation

(b.) t = [itex]\displaystyle \frac{v_{0}sin\alpha_{0}}{g}[/itex], where g is the acceleration due to gravity.
I would call this t1, or tUP or some other label of your liking.
This is obtained by using the kinematic equation: v = v0y + at, plugging in the vertical component for the initial velocity

(c.) [itex]\displaystyle \frac{(v_{0}sin\alpha_{0})^{2}}{2g}[/itex] + H = Maximum height above ground
You could have also used h - H = (1/2)gt1, since the time it take to achieve max. height is equal to the time from max height to height, H. (getting the same answer.)
I obtained this the same way I obtained (b.) only this time, I used a different kinematic equation: v2 = [itex]v^{2}_{0}[/itex] + 2a[itex]\Delta[/itex]r

The added H should take into account the missing height from the ground to the launch height.

So if the launch height was equal to ground level, I could just take the time it took to reach the maximum height since the motion should be symmetrical about the max height. Could I use that information somehow?
Yes, What is the vertical component of the projectile when it gets back to an elevation of H? Find the time from H (on the way down) + twice t1.

Otherwise, use (1/2)gt22 to find the time from max height to the ground.
 
  • #5
Bandersnatch
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Don't you want to know when the position of the body equals 0 on the y(height) axis? Perhaps using appropriate kinematic equation would help.
 
  • #6
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What is the vertical component of the projectile when it gets back to an elevation of H? Find the time from H (on the way down) + twice t1.
The vertical component would be -[itex]v_{0}[/itex]sin[itex]\alpha_{0}[/itex], and dividing this by the acceleration due to gravity g, would present me with the time to reach the elevation H.

Don't you want to know when the position of the body equals 0 on the y(height) axis? Perhaps using appropriate kinematic equation would help.
Alright, I think I'm gettin' this now.

Using the following kinematic equation: [itex]\Delta[/itex][itex]r_{y}[/itex] = [itex]v_{0}[/itex]t + [itex]\frac{1}{2}[/itex]a[itex]t^{2}[/itex]

Plugging in known vertical components, and solving for time, I get: [itex]\sqrt{\frac{2H}{g}}[/itex] = t, where t is the time from height H to the ground. Combining this equation with the above, I get:

[itex]\frac{v_{0}sin\alpha_{0}}{g}[/itex] + [itex]\sqrt{\frac{2H}{g}}[/itex] = t, where t is the time from the peak of the motion to the ground below height H

This look right?
 
  • #7
SammyS
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The vertical component would be -[itex]v_{0}[/itex]sin[itex]\alpha_{0}[/itex], and dividing this by the acceleration due to gravity g, would present me with the time to reach the elevation H.

Alright, I think I'm gettin' this now.

Using the following kinematic equation: [itex]\Delta[/itex][itex]r_{y}[/itex] = [itex]v_{0}[/itex]t + [itex]\frac{1}{2}[/itex]a[itex]t^{2}[/itex]

Plugging in known vertical components, and solving for time, I get: [itex]\sqrt{\frac{2H}{g}}[/itex] = t, where t is the time from height H to the ground. Combining this equation with the above, I get:

[itex]\frac{v_{0}sin\alpha_{0}}{g}[/itex] + [itex]\sqrt{\frac{2H}{g}}[/itex] = t, where t is the time from the peak of the motion to the ground below height H

This look right?
No.

[itex]\displaystyle \sqrt{\frac{2H}{g}}=t[/itex] is the time to fall to the ground starting at rest from height H.

Either start at rest from max height: [itex]\displaystyle h_{max}=\frac{(v_{0}sin\alpha_{0})^{2}}{2g}+H[/itex]

or,

Start at height H with initial downward component of [itex]-v_0\sin(\alpha_0)\ .[/itex] To get back down to H takes a time of [itex]2\,t_1\ [/itex] after launch.
 
  • #8
Bandersnatch
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If I may, I've always found it more comprehensible to take the position equation
[itex]r = \frac{1}{2}at^{2} + v_{0}t + r_{0}[/itex]
plug in the vertical components only, and look for roots.
Just disregard the non-physical solution.
 
  • #9
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I apologize for the late response, but I don't want to leave this unresolved.

I did use the above advice, using [itex]\displaystyle h_{max}=\frac{(v_{0}sin\alpha_{0})^{2}}{2g}+H[/itex] as my value for the initial position in the vertical plane. And I used the kinematic equation: [itex]r = \frac{1}{2}at^{2} + v_{0}t + r_{0}[/itex].

I then received this as my equation for the position on the ground in relation to its position at the peak of its motion:


[itex]v_{0}t - \frac{1}{2}at^{2} + \frac{(v_{0}sin\alpha_{0})^{2}}{2g}+H[/itex]

The second term in that equation is negative because it's still just the acceleration due to gravity. With that in mind, I can solve for the time to reach the ground from the peak. The initial velocity at the top should also be zero, so that makes this equation much simpler to deal with (I think). In any case, I received the following result:

t = [itex]\sqrt{\frac{\frac{(v_{0}sin\alpha_{0})^{2}}{g}+2H}{g}}[/itex]

This is also dimensionally correct, so I'm hoping this is correct
 
  • #10
Bandersnatch
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The beauty of the position equation, is that you don't even have to take the hmax as your starting point. That equation describes the projectile's position at every point of its trajectory as a function of time.
So you could just as well start at the lauch point, with H, V0sinα and -g as the three parameters in the vertical direction, and get your roots from there.
The end result should be the same, but it's so much simpler not having to worry about cutting the trajectory into pieces and then reassembling the results from each piece and so on.
 
  • #11
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The beauty of the position equation, is that you don't even have to take the hmax as your starting point.
Ah that's pretty neat!
 

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