Projectile Angle Calculation: Finding Optimal Launch Angles for Accurate Hits

[ericka141]

Homework Statement

You shoot a cannon at a distant cliff. The cliff is 170m high and 550m away, with the launch speed of the projectile being 110 m/s. Wind resistance is ignored. At what angles, relative to the horizontal, should the projectile be shot to hit the target?

Hint: 1/cos2θ = 1+tan2θ

Homework Equations

Kinematics Equations:
vf=vi + a*t
d= vi*t + 1/2a*t^2
d= (vf^2-vi^2)/2a

The Attempt at a Solution

I've broken the problem up into x and y components:
vix= vixcos(θ)
dx= 550m (or vixcos(θ)*t)
ax= 0m/s^2

viy= viysin(θ)
dy= 170m (or viysin(θ)*t)
ay= -9.81m/s^2

I know how to find the maximum distance that a projectile travels when given the initial velocity (or the distance when given the initial velocity and the angle it's shot at) but I'm not sure how to find these angles. I'm not exactly sure where tan(θ) comes in. If somebody could help me with this that would be GREAT, this assignment is due tonight.

 

[Bryson]

If you know x and y (or $v_x$ and $v_y$), you can find the angle. Who is the man? Tan is the Man!

Recall SOH CAH TOA

 

[ericka141]

Then how would I use 1/cos2θ=1+tan2θ ??

 

[Bryson]

You don't. $tan(\theta) = \frac{d_x}{d_y}$

Solve for $\theta$

 

[Nickie]

I would approach this problem by first understanding the basic principles of projectile motion and the factors that affect it. From the given information, we know that the projectile has a constant initial velocity of 110 m/s and is affected by gravity (ay = -9.81 m/s^2). We also know that wind resistance is ignored.

To find the optimal launch angles for accurate hits, we need to consider the horizontal and vertical components of the projectile's motion. As you have correctly identified, the horizontal component (x) is affected only by the initial velocity (vix) and time (t) and is given by dx = vix * t. On the other hand, the vertical component (y) is affected by both the initial velocity (viy) and the acceleration due to gravity (ay), and is given by dy = viy * t + 1/2 * ay * t^2.

Now, to find the optimal launch angles, we need to consider two things: the maximum distance traveled by the projectile (which would result in hitting the target) and the angle at which this maximum distance occurs. Using the kinematics equations, we can calculate the maximum distance (dmax) as:

dmax = (vix^2 * sin(2θ)) / g

where g is the acceleration due to gravity, 9.81 m/s^2.

To find the angle at which this maximum distance occurs, we can use the hint provided: 1/cos2θ = 1 + tan2θ. This can be rearranged to solve for tan2θ, which will give us the angle at which the maximum distance occurs. Once we have this angle, we can solve for θ by taking the inverse tangent (tan^-1) of the result.

In summary, to find the optimal launch angles for accurate hits, we need to use the kinematics equations to calculate the maximum distance and then use the hint provided to solve for the angle at which this maximum distance occurs. I hope this helps and good luck with your assignment!