Projectile motion problem involving finding launch angleby EddieHimself Tags: angle, involving, launch, motion, projectile 

#1
Jun2811, 11:33 AM

P: 9

1. OK so if you imagine a projectile is fired with an inital velocity (u) of 400 m/s at an angle theta. What two values of theta will cause the projectile to hit a point b that is 5 km horizontally away from the point of launch and 1.5 km vertically?
u (magnitude)= 400 m/s deltaH = 1500 m (1.5 km) deltaS = 5000 m (5 km) 2. constant acceleration equations, trigonometric identities. 3. What i tried to do was: u_{x}=400cos(theta) u_{y}=400sin(theta) using s=ut+1/2at^{2} 1500 = 400t.Sin(theta)4.905t^{2} (1) 5000 = 400t.cos(theta) t = 12.5/cos(theta) (2) substituting (2) into (1) gives, 1500 = 5000Sin(theta)/Cos(theta)  766.4/Cos^{2}(theta) multiplying by Cos^{2}(theta) gives 1500cos^{2}(theta) = 5000sin(theta)cos(theta)  766.4 = 2500Sin(2theta)  766.4 = 5000Cos^{2}(theta)  2500  766.4 3500cos^{2}(theta) = 3276 cos^{2}(theta) = 0.9361 cos(theta) = 0.9675 theta = 14.6° which is the wrong answer. Obviously i'm missing something here, but i don't know what. Hence why i'm posing the question on here. Thanks EH. 



#2
Jun2811, 12:36 PM

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P: 11,446





#3
Jun2811, 12:48 PM

P: 9





#4
Jun2811, 01:25 PM

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P: 11,446

Projectile motion problem involving finding launch angleso it's not a valid identity. 



#5
Jun2811, 01:55 PM

P: 9

Ok i've sorted this one now;
sodding the cos^{2}(theta) division, i found that i could instead use the substitution sec^{2}(a) = 1 + tan^{2}(a) and then what resulted was 1500 = 5000tan(theta) 766.4tan^{2}(theta)  766.4 which rearranged to a simple quadratic, that being; 766.4tan^{2}(theta)  5000tan(theta) + 2276 = 0 using the equation; tan(theta) = 0.49 or 6.034 that gives the 2 angles of theta being 26.1° and 80.6° which matches up with the answers in the book so that's fine. 



#6
Jun2811, 02:02 PM

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P: 11,446

Yup. That'll do it!
Another approach would be to abandon the trig functions altogether and use vx and vy as the velocity components, with the additional equation v^{2} = vx^{2} + vy^{2}. Solve the three equations for either vx or vy (since the other can be had by Pythagoras, and the angle is trivial given vx and vy). The resulting equation in one variable (say vx) looks like a quartic, but since there's only vx^{4} and vx^{2} in it, it's really a quadratic at heart. 


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