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

  • Thread starter LexRunner
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
Dnomyar
Gold Member
29
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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/
 
Last edited:
  • #3
tjmiller88
Gold Member
13
2
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
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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
gneill
Mentor
20,925
2,866
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...
 
Last edited:

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