basic projectile Physics 101 problem......

[Squid]

Gun fires at 500m/sec at target 800m down range.

What angle does gun need to be fired at to hit target?

No air resistance, just bonehead physics problem.

I get that height(y) related to distance with Cos and Sine related to gravity, but how to set it up and why?

Basically, the angle up will be just enough to keep bullet launch and impact at same height for time bullet takes to travel to target.

Once bullet leaves gun, its HORIZONTAL (in this problem) will be constant, but wont be full 500m due to angle.

Any help?

Please don't junk up this thread with cute comments. Respond only if you know or have link to good explanation. Actual homework in progress. I'll be back in a bit.

[Peter.Steele]

First link should cover you ... All the equations are there.

[Thefeeder]

Gun fires at 500m/sec at target 800m down range.

What angle does gun need to be fired at to hit target?

No air resistance, just bonehead physics problem.

I get that height(y) related to distance with Cos and Sine related to gravity, but how to set it up and why?

Basically, the angle up will be just enough to keep bullet launch and impact at same height for time bullet takes to travel to target.

Once bullet leaves gun, its HORIZONTAL (in this problem) will be constant, but wont be full 500m due to angle.

Any help?

Please don't junk up this thread with cute comments. Respond only if you know or have link to good explanation. Actual homework in progress. I'll be back in a bit.

At 500m/sec...how long will the bullet take to travel 800m.... call it X


How far will an object fall in X seconds. (in vacume)

[CSACANNONEER]

If this is for a class, the real answer is that not enough info has been given. Sure, one can assume that both the shooter and target are at the same elevation but, how can one be positive that this will always be the case?

[Squid]

read it again Feeder, it wont be going at target at 500m/s because it needs to

angle up slightly, hence the riddle.


CSACANNONEER, enough is given as I stated this is bonehead physics problem and everything (gun and target) is level, perfectly mathimatically.


"....Basically, the angle up will be just enough to keep bullet launch and impact at same height for time bullet takes to travel to target......"


There is a plugin equation on the wiki sight but we are supposed to do it with 'unit notation' and trig and acceleration of gravity, like we know what we are doing.

[russ69]

Acceleration of gravity and time of flight should give you the elevation needed to find the angle.

[G-forceJunkie]

I dont know the formula but figure out the flight time, figure out how much the bullet will drop in that amount of time with the constant of gravity. Once you have the distance and drop, you can plug them into a right triangle formula to get the angle. Not having my books in front of me its something like tanA= a/b where a is the drop and b is the 800m.

[G-forceJunkie]

Ok, you made me pull out the caculator.

1.6 sec bullet flight
9.8 meters per second gravity acceleration
15.7 meters drop over 800 meters

tanA=a/b
a=15.7
b=800
angle A = 1.12 deg

[HighLander51]

Acceleration due to gravity is meters/second^2 because acceleration is the first derivative of velocity.

Here is the simple ballistic model.

http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

and here is a plug and chug version

http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=f630255c

The iphone has an app also.

[G-forceJunkie]

But both of those take in to account the bullet slowing down due to air resistance, something not part of the OP's math problem.

Acceleration due to gravity is meters/second^2 because acceleration is the first derivative of velocity.

Here is the simple ballistic model.

http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

and here is a plug and chug version

http://www.shooterscalculator.com/ballistic-trajectory-chart.php?t=f630255c

The iphone has an app also.

[lilro]

Don't you need to know the weight of the projectile to solve this?

[MMA]

Don't you need to know the weight of the projectile to solve this?

probably not:

[russ69]

Don't you need to know the weight of the projectile to solve this?

Law of falling bodies. Weight does not affect acceration due to gravity.

[Yemff]

Maybe it's just me but it seems you need some other piece of info to solve this. You don't have the x or y components of the velocity and you don't have the the angle so I don't see how you can solve for one without the other. Now if you had either the time the projectile is in the air, or the height it reaches you would be able to solve for either the x or y components of the velocity, but you have neither. Maybe i'm missing something here, but I don't think you can solve it with the information given.

In response to some answers given above, 1.6s can not be the time the projectile is in the air. You are assuming the horizontal component of the velocity is 500m/s, but since the bullet is fired at an angle, this x component of velocity will be less than 500m/s.

[Squid]

Hint, I believe there will be two answers......

1) for 'typical' trajectory

2) for "high angle"


Common sense says "yes, we have enough info, because we sorta know there is a certain particular angle that would make it hit".