Finding Optimal Angle to Hit Target 600m Away with 30m Hedge

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically calculating the optimal angle to launch a cannonball over a 30-meter-high hedge to hit a target located 600 meters away. The initial speed of the cannonball is given as 100 meters per second, and participants are exploring the necessary conditions to ensure the projectile clears the hedge while reaching the target.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to determine the angle required for the cannonball to clear the hedge and reach the target. Questions arise regarding the distance from the cannon to the hedge and the implications of different launch angles on the projectile's trajectory.

Discussion Status

There is an ongoing exploration of various angles and their effects on the projectile's path. Some participants have provided hints and suggestions for calculating the angle, while others express confusion about the relationship between the angle, distance, and height. Multiple interpretations of the problem are being discussed, particularly regarding the requirements to clear the hedge and hit the target.

Contextual Notes

Participants are working under the constraints of the problem statement, which specifies the height of the hedge and the distance to the target. There is a lack of consensus on the correct angle, with some suggesting that a 45-degree angle is sufficient, while others argue for a different angle based on the need to clear the hedge and hit the target accurately.

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I'm trying to solve a "riddle" which is a bout shooting a cannonball above a hedge with a heigth of 30 meters. My goal is to hit a target on the other side of the hedge, at least 600 meters from the point where the cannon is located. The speed of the cannonball is approximately 100meters/second. The weight of the cannonball is irrelevant.

I'm looking the optimal angle and distance. Can't move closer than 600.

So I have length = 600m or further
Speed leaving cannon = 100m/s
Cannonball has to gove above the heigth 30m.

And I´m looking for the best angle. Or just an angle that will do the job..

It's been a long time since I read about physics, so I'd be very thankful if someone could freshen my memory a bit. I don't remember the formula for these type of calculations.
 
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What is the distance from the canon to the hedge?
 


nure said:
What is the distance from the canon to the hedge?

Sorry I missed that part, the distance from cannon to the hedge is 500 m. So the target behind on the other side of the hedge is 100 m behind the hedge.
 


If the initial speed makes an angle with the horizontal larger than 39.4 degrees, the cannonball will go over the hedge 500 meters away. And as we know the ball will have the maximum reach when the angle is 45.

But then the cannonball will not hit the target 100 meters behind the hedge, it will go further.
 


Could you please explain how you acquired that result?
Thanks a lot.
 


Hints:

1. Determine the angle for the shot to go 600 m.
2. Based on this angle, determine the height above ground when the ball reaches the hedge. If it is less than 30 m, you do not have the correct solution.
3. Look for another angle where the trigonometry function has the same value. Think mortar shot rather than a cannon shot if the ball won't clear the hedge.
 


Still don't get it right :/
 


The angle for the shot to go 600 m is not 39.4 degrees.
 


I'd be thankful if you'd show me the correct solution.
 
  • #10


I cannot do it for you but I'll be glad to help. First thing to do is to determine the angle that will allow a 100 m/sec shot to travel 600 m. Let's forget about the hedge for the moment.

To do this you assume that the horizontal velocity is constant. This is true if drag is neglected which is our case here. Determine an expression for the range by finding the horizontal component of velocity and multiplying it by the time of flight. It'll be a function of angle and time of flight. You do not know either.

Next thing to do is determine the time of flight as a function of the vertical component of initial velocity. It will be a function of the unknown angle also. Get the expression for time and eliminate the unknown time from the range equation. Don't forget to double it. That leaves one unknown, namely theta.

Set the range equation equal to 600 m and solve for the angle.

Then test to see if ball clears the hedge. Get back to me when you have done this.
 
  • #11


LawrenceC said:
The angle for the shot to go 600 m is not 39.4 degrees.

The angle of 39.4 degreed will though be the smallest angle the initial speed can make with the horizontal to get the cannonball not to collide with the hedge.
 
  • #12


You don't want it to hit the hedge. You want it to clear the hedge and impact the target that is 600 m from where it was shot.
 
  • #13


LawrenceC said:
You don't want it to hit the hedge. You want it to clear the hedge and impact the target that is 600 m from where it was shot.

Well, you won't hit the hedge if the angle is greater than 39.4? And the main question is asking for an angle that will get the cannonball AT LEAST 600 meters away from the initial position. If he actually want to hit the ground 600 meters ayaw, surely the angle will be different.
 
  • #14


Here is the problem statement: "I'm trying to solve a "riddle" which is a bout shooting a cannonball above a hedge with a heigth of 30 meters. My goal is to hit a target on the other side of the hedge, at least 600 meters from the point where the cannon is located."

Problem says hit the target. You can clear the hedge as well as hit the target.
 
  • #15


I've solved the problem using a 45 degree angle. Simple but sufficient for me. Thank you for the hints!
 
  • #16


If you shoot it at 45 degrees, the ball lands about 1019 meters from where it was shot. You've missed the target by over 400 meters.

The correct answer is 71.97 degrees.
 

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