How to find angle of launcher in order to hit target?

In summary, the problem is to find the angle of a marble launcher that is 1.17 m tall and needs to hit a target 3.645 m away. The marble is launched at a height of 1.17 m and the question asks how to solve this without knowing the velocity. The conversation suggests using various equations and resources to find the angle and velocity, but there may be missing information from the teacher. The student also mentions a previous lab where they launched a marble horizontally from a height of 3.645 m, but the calculations seem off and there may be confusion about the numbers used. In short, more information and clarification is needed to accurately solve the problem.
  • #1
LexRunner
7
0
This is the problem:
I have a marble launcher that is 1.17 m tall and it needs to hit a target that is 3.645 m away. What should the angle of the launcher be set to in order to hit the target?
*Marble is launched at a height of 1.17 m.

What I don't understand:
How can I solve this without the velocity? Am I suppose to find the velocity? And how do I find it?
After I do find the velocity, how am do I find time? Do I plug the info into this equation: d=vit + 0.5at2 to find the time? Then I am completely lost because I don't know what to do to find the angle.

Please I need help. It is 12:30 AM and I have a graded lab on this tomorrow.
 
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  • #2
http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

for various formulae, but I couldn't see the one you were using.

You are right, you need to specify the velocity - it may be part of the trick, in the question.

I would simply supply the angle and velocity to reach the target ( there would be many options ). As long as you present the calculations accurately, they couldn't mark you wrong whatever velocity you used .


Check your answers at
http://www.calctool.org/CALC/phys/Newtonian/projectile

Good explanation here
http://www.instructables.com/answers/How-to-calculate-trajectory-of-projectile/
 
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  • #3
Hi LexRunner,

Yea it looks like you might be missing some info from your teacher.

What I would do is assign the velocity a variable called v0 to the initial velocity, and then solve the problem in terms of that. Then you can ask you teacher what that should be, and then plug in if needed.

Okay so your first step - you need to draw out the problem and label your x and y coordinates. Then label your angle θ. You have your target where you know x=3.645m and y=0 (I'm assuming).

From this point you will need to establish you x and y position equations in terms of t.

Once you have those, you know there is some time ttarget where x=3.645m and y=0. Based on that information you will be able to solve for θ in terms of v0.

A little bit hard to explain without a diagram - so if you're still stuck, PM me and I'll send you a link to a similar example problem I posted a while back. Hope this helps.
 
  • #4
We actually did a lab a few days before where we did a zero degree launch. We set the marble launcher on top of the lab tables and launched the marble at 0 degree (horizontal launch). When I launched the marble, the x distance it traveled was 1.17m and the y height the marble was launched at was 3.645m. I did some calculations and found the time to be 4.88s and then found the velocity to 7.469m/s. Can someone check to see if the work I did was correct? And if it is, can I plug that velocity into my original problem to find the angle of launch?
 
  • #5
LexRunner said:
We actually did a lab a few days before where we did a zero degree launch. We set the marble launcher on top of the lab tables and launched the marble at 0 degree (horizontal launch). When I launched the marble, the x distance it traveled was 1.17m and the y height the marble was launched at was 3.645m. I did some calculations and found the time to be 4.88s and then found the velocity to 7.469m/s. Can someone check to see if the work I did was correct? And if it is, can I plug that velocity into my original problem to find the angle of launch?

The numbers don't look right. A nearly 5 second "hang time" from a height of about 4 m should have triggered a suspicion!

Can you show your work?

EDIT: The numbers 1.17 m and 3.645 m show up twice in different contexts in this thread. In the first case 3.645 m was the distance to a target, in the latter case it was the height of the horizontal launch (which seems rather odd because, hey, who has a lab bench next to a 4 m high cliff?) . You first stated that 1.17 m is the launch height for your launcher, then you say 1.17 m is the horizontal distance traveled when the marble is launched horizontally from a height of 3.645 m. The repetition of the same numbers seems an unlikely coincidence.

I think you need to sort through your data and clearly identify which numbers are which...
 
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1. How do you calculate the angle of a launcher to hit a target?

The angle of a launcher can be calculated using the formula: θ = tan^-1 (h/d), where θ is the angle, h is the height of the target, and d is the distance to the target.

2. What factors should be considered when determining the angle of a launcher?

Factors that should be considered include the velocity and trajectory of the projectile, the height and distance of the target, and any external forces such as wind or gravity.

3. Is there a specific angle that will guarantee hitting the target?

No, there is no specific angle that will guarantee hitting the target. The angle will depend on the specific variables and conditions of the launch.

4. How can you improve the accuracy of the angle calculation?

To improve accuracy, it is important to measure and input precise values for the variables, including the height and distance of the target, and to account for any potential external forces.

5. Are there any tools or techniques to help determine the angle of a launcher?

Yes, there are various tools and techniques such as using a protractor or inclinometer, utilizing mathematical formulas, or using computer simulations to help determine the angle of a launcher.

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