Find angle of release

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  • #1
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A cannon that shoots projectiles at 100m/s must hit a target 400m away and 30m above the ground. What angle must the cannon point at in order to hit the target?
I have two functions:
x = V*cos(theta)*t, where V = initial velocity, and
y=V*sin(theta)*t - (g*t^2)/2 where g = gravity's acceleration
but i can't solve for theta.
can somebody help me?
 

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  • #2
HallsofIvy
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Why can't you solve for theta?

Since you are trying to hit a target 400 m away with initial speed 100 m/s, the horizontal equation, x = V*cos(theta)*t, becomes
400= 100 cos(theta)t or t= 40/cos(theta).

Since the height of the target is 30 m above the ground (which, I assume, is the height of the cannon), the vertical equation,
y=V*sin(theta)*t - (g*t^2)/2, becomes 30= 100 sin(theta) t- (g t<sup>2</sup>)/2. Now using t= 40/cos(theta), we can write that as 30= 4000 (sin(theta)/cos(theta))- 800g/cos<sup>2</sup>(theta).

Multiplying both sides by cos<sup>2</sup>(theta),
30 cos<sup>2</sup>(theta)= 4000 sin(theta)cos(theta)- 800g

That's a little complicated but can be solved.
 
  • #3
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thanks

cool i solved it, it was pretty damn long
 
  • #4
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hmm

great :) this kinda answeres the first part of my question earlier...
 
  • #5
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er

:| just one question... the variable t in that equation is time... correct?
 
  • #7
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hmm

ok... well this kind of explains a few things about designing the trajectory of a bullet fired from a weapon at a given angle. let me see if i have this right.
If a bullet is fired at a specific angle and muzzle velocity then it would be at point x and y which can be derived from the equations at time t right?

like if V = 1200 m/s and angle of fire is 60* then the bullet would be at point (12000, 18824) right?
 
  • #8
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The solutions in the beginnins solve different stuff including t i think.

in the last solution, if the bullet initial velocity is 1000m/s and the target is at point (3000, 4000) then, the equation solves the firing angle.

Its a little hard to explain what it does but here's a tip. read it over 3-4 times and ull prob get it.

so basically it solves for theta in the equations x = (Vx)t, and
y = (Vy)t + (1/2)at^2

so first it solves algebraicaly for t in the first equation which becomes t = x / (Vx), which is 3000m/Vx or 3000m/(1000cos(theta))

then it substitutes it for t in the second equation in the third line of the solution.
then it reduces it in the fourth line, then multiplies both sides by cos^2(theta)
then it reduces again which results in the fifth line and then
substitutes sin(theta) for sqrt(1 - cos^2(theta))
then after squaring both sides and reducing the equation becomes a second degree one like ax^2 + bx + c = 0 , where x is cos^2(theta)
then it uses the quadric formula
x = ( -b -+ sqrt(b^2 - 4ac) ) / (2a) to find cos^2(theta) then from there its easy it find the square root of the answer and then finds the cos-1 of the answer to find theta;
lol thats long, a friend told me theres a much easier way in advanced physics school books.
 
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