Projectile motion problem involving finding launch angle

  • #1
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:

ux=400cos(theta)
uy=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.
 

Answers and Replies

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

How'd you get from (A) to (B)?
 
  • #3
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.
 
Last edited:
  • #4
gneill
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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.
 
  • #5
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.
 
  • #6
gneill
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Yup. That'll do it! :smile:

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.
 

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