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Projectile Problem - Arrow and Target

  • #1

Homework Statement


Target #1 and Target #2 are placed as shown. The one on top is made of paper so that an object hitting it will pass right through without its motion being disturbed. At what velocity and angle must you shoot an arrow so that it will pass through target #1 and land on target #2? Assume that the launch and land heights are the same.

[URL=http://img296.imageshack.us/my.php?image=21453445op0.jpg][PLAIN]http://img296.imageshack.us/img296/232/21453445op0.th.jpg[/URL]http://g.imageshack.us/thpix.php [Broken][/PLAIN]


Homework Equations


Range equation is Vo^2 sin 2θ / g
horizontal velocity = Vo cos θ
vertical velocity - Vo sin θ

That's all I can think of!

The Attempt at a Solution


I'm thinking that the horizontal velocity is the range, but I can't even get the range since I don't have Vo and θ. I usually find vertical and horizontal components, but I don't have Vo and θ. Is the distance from the arrow to target #2 (100 m) the range, since it's X? I'm really not sure how to approach this problem, and projectile motion isn't my cup of tea...
Would you use trig to find horizontal velocity since you have distance half the trip in the air and the height from the first target to the floor?
 
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Answers and Replies

  • #2
Hootenanny
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HINTS:

(a) What is the required maximum height so that the arrow hits the first target?

(b) What is the required range so that the arrow hits the second target?

Don't worry you won't actually be able to calculate them explicitly yet, I'm just looking for equations.
 
  • #3
Ummm
Req maximum height to hit the first target is 40 m. I know that at target 1 that Vy is 0, right?
The required range so that the arrow hits target 2 is 100 m.

Vy^2 = Vyo^2 + 2a(y-yo) ???
 
  • #4
Hootenanny
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Ummm
Req maximum height to hit the first target is 40 m. I know that at target 1 that Vy is 0, right?
The required range so that the arrow hits target 2 is 100 m.

Vy^2 = Vyo^2 + 2a(y-yo) ???
Correct.

So what is the required initial vertical velocity for the projectile to reach a maximum height of 40m?

How long is the particle in the air? Hence, what would be the required horizontal velocity?
 
  • #5
Vy is zero and I assumed Yo to be 0 too. I got Vyo to be 28 m/s.

The only equation I know that relates all my info given is

y= Yo + VyoT + 1/2 at^2

to get time. Am I supposed to use that?

I don't know about horizontal velocity though. Trig?
 
  • #6
Hootenanny
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Vy is zero and I assumed Yo to be 0 too. I got Vyo to be 28 m/s.

The only equation I know that relates all my info given is

y= Yo + VyoT + 1/2 at^2

to get time. Am I supposed to use that?
Good, you can use that equation to determine the flight time, i.e. the time from launch to landing.
I don't know about horizontal velocity though. Trig?
Once you have determined the flight time, you should be able to find the horizontal velocity trivially.
 
  • #7
so I solved for time and got 2.85 seconds

using Vx = x/t
I get Vx to be 17.5 m/s.
 
  • #8
ShawnD
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Vy is zero and I assumed Yo to be 0 too. I got Vyo to be 28 m/s
[......]
so I solved for time and got 2.85 seconds

using Vx = x/t
I get Vx to be 17.5 m/s.
I got those as well, so I guess that's a good sign.

Now you can use those to find the angle since:
tan(angle) = y/x (remember your unit circle), so then:
tan-1 (y/x) = angle

I get the angle as 58 degrees. Finding the initial velocity from that should be easy :)
 
  • #9
Yes so I got Vo as 33 m/s
and I did get theta as 58!

Thank you so MUCH!
 

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