# Projectile motion problem involving finding launch angle

• EddieHimself
In summary: So if you're trying to solve for vx or vy, you can ignore the quartic part and just solve for vx2 and vy2. In summary, the projectile will hit a point b 5 km horizontally away from the point of launch and 1.5 km vertically. If a = 0, then sin(2a) = 0, while 2cos2(a) - 1 = 1. Therefore, the trigonometric identity is not valid. If you abandon the trig functions, vx and vy can be solved for using vx2 and vy2.
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:

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.

EddieHimself said:
(A)... = 2500Sin(2theta) - 766.4
(B)... = 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!

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.

## 1. What is projectile motion?

Projectile motion is the motion of an object that is projected into the air and then allowed to move under the influence of gravity.

## 2. How do you find the launch angle in a projectile motion problem?

The launch angle can be found by using the equation tanθ = vy/vx, where θ is the launch angle, vy is the vertical velocity, and vx is the horizontal velocity.

## 3. What is the initial velocity in projectile motion?

The initial velocity in projectile motion is the velocity with which the object is launched.

## 4. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object and changing its trajectory. However, it is often neglected in projectile motion problems unless specified otherwise.

## 5. Can the launch angle affect the range of a projectile?

Yes, the launch angle can affect the range of a projectile. The optimal launch angle for maximum range is 45 degrees, but the range can be affected by other factors such as air resistance and initial velocity.

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