Projectile motion problem involving finding launch angle

[EddieHimself]

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+1/2at²

1500 = 400t.Sin(theta)-4.905t² (1)

5000 = 400t.cos(theta)

t = 12.5/cos(theta) (2)

substituting (2) into (1) gives,

1500 = 5000Sin(theta)/Cos(theta) - 766.4/Cos²(theta)

multiplying by Cos²(theta) gives

1500cos²(theta) = 5000sin(theta)cos(theta) - 766.4

= 2500Sin(2theta) - 766.4
= 5000Cos²(theta) - 2500 - 766.4

3500cos²(theta) = 3276

cos²(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.

 

[gneill]

(A)... = 2500Sin(2theta) - 766.4
(B)... = 5000Cos²(theta) - 2500 - 766.4

How'd you get from (A) to (B)?

 

[EddieHimself]

How'd you get from (A) to (B)?

Thanks for pointing that out, yes it was actually Cos2a and not Sin2a I just read it wrong, probably a bit tired lol. Ok thanks for your help.

 

[gneill]

Trigonometric identity; Sin(2a) = 2Cos²(a) - 1

Suppose a = 0. Then sin(2a) = 0, while 2cos²(a) - 1 = 1.

so it's not a valid identity.

 

[EddieHimself]

Ok I've sorted this one now;

sodding the cos²(theta) division, i found that i could instead use the substitution sec²(a) = 1 + tan²(a)

and then what resulted was

1500 = 5000tan(theta) -766.4tan²(theta) - 766.4

which rearranged to a simple quadratic, that being;

766.4tan²(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.

 

[gneill]

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 v² = vx² + vy². 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 vx⁴ and vx² in it, it's really a quadratic at heart.