Projectile motion problem involving finding launch angle

AI Thread Summary
A projectile launched at an initial velocity of 400 m/s must hit a target 5 km horizontally and 1.5 km vertically, requiring the calculation of two launch angles. The user initially attempted to derive the angles using trigonometric identities and constant acceleration equations but encountered errors in their calculations. After correcting a mistake regarding the use of sine and cosine, they successfully formulated a quadratic equation in terms of the tangent of the angle, yielding two valid angles of 26.1° and 80.6°. An alternative method suggested involves using velocity components without trigonometric functions, simplifying the problem to a quadratic equation. The discussion highlights the importance of careful application of trigonometric identities in projectile motion problems.
EddieHimself
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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.
 
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EddieHimself said:
(A)...[/color] = 2500Sin(2theta) - 766.4
(B)...[/color] = 5000Cos2(theta) - 2500 - 766.4

How'd you get from (A) to (B)?
 
gneill said:
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:
EddieHimself said:
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.
 
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.
 
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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