# Projectile pitch

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If a projectile is travelling at say 6mph and I know it covers 2miles before hitting it's target, how can I work out what angle it was fired at? I am looking for the angle of pitch about the x axis, anyone know how this is worked out?

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Let the angle be U

You know the initial velocity V
You know the range, R = Vx * t = Vcos(U)*t
You know the acceleration due to gravity, ~9.8m/s^2

Since the y velocity goes from +Vy to 0 and then back down, the time spent in the air is the time required to change the vertical velocity by 2Vy at a rate of g

t = 2Vy / g = 2Vsin(U) / g

R = Vcos(U) * (2Vsin(U) / g)
= 2 V*V * cos(U) * sin(U) / g
= V*V* sin(2U)/g

sin(2U) = g * R / (V * V)

Now you know the sine of twice the angle, you can take the inverse sine, and halve that to get your answer.

Also, this is a programming forum, not a physics forum.

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I have to make a couple assumptions: 1) the projectile was fired from the same elevation at which it lands, and 2) the speed you mention (6 mph) is the horizontal speed. It's vertical speed will vary as it travels the parabola from firing to landing (which you seem to understand).

This is all off the top of my head ( <-- excuse if it's wrong ).

The transit time from firing to landing:
tTime = horizontal_distance/horizontal_speed;

Because I assumed it is fired and lands at the same elevation, then it reaches it's peak height (with a velocity of zero) at the halfway point, or tTime/2 seconds from the moment it was fired.

The projectile is subject to gravity and the law distance = 1/2 * acceleration * time-squared. More importantly, the vertical velocity is acceleration * time.

The acceleration due to gravity is 32 feet/(second*second). The time is tTime/2.

So, the projectile's initial vertical velocity (and the vertical velocity when it lands) is 32 feet/(second*second) * tTime/2, or 16*tTime feet/sec.

When it's fired, the angle with the horizontal is atan(vert_vel/horiz_vel).

To convert from mph to feet/sec: feet/sec = mph * 5280 ft/mile * 1hour/3600 seconds - or feet/sec = mph * 1.4667.

Angle = atan( 16*tTime / (mph *1.4667) );

With your numbers, tTime = 2 miles/6 mph = 20 minutes, or 1200 seconds.

The vertical velocity at firing = 16*1200 = 19200 feet/second or approximately mach 25. That is, air friction melted it before it left the barrel.

In any case, angle = atan(19200/8.8) = 89.97 degrees.

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More likely, he was given the initial velocity of the projectile. In that case, the given numbers are impossible.

If you shoot something at 6mph, the maximum possible range is 73 centimeters!

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Quote:
 Original post by gameplayprogammerIf a projectile is travelling at say 6mph and I know it covers 2miles before hitting it's target
If it's travelling 6mph, but it goes for 2 miles, that means it would take 20 minutes to arrive if it was travelling in a straight line... I dare say your target would have moved by then :)

Also, I would probably say that it's much easier when working in metric to work out things like this because all of the forumlas work with SI units (metres, seconds, kilograms, etc)

But anyway, storyyeller's forumla will be able to tell you the answer once you convert the units to SI (and fix up the "2 miles at 6mph" issue)

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You don't have to convert the units at all if you have a calculator that can handle unit conversion automatically.

It's still a good idea in general though.

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Quote:
 Original post by StoryyellerYou don't have to convert the units at all if you have a calculator that can handle unit conversion automatically.
Wow, I've never seen one of those! I guess coming from Australia, we've never had the need for such things since metric is the "natural" way for us to think :)

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Well I was thinking of Google when I said that.

For example, you can do something like
((6mph)^2 / (9.8m/s^2)) and it automatically handles all the conversions. (The answer is 73.4123614 cm or 0.000456163265 miles)

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Moving to Math and Physics. There's no actual programming content here (yet, anyway).

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Thanks for the replies, the 6mph is meant to be 60mph btw sorry :-)

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