Angular Velocity: Find Launch Angles for Projectile

In summary, the problem is asking for the two angles at which a projectile must be launched at 36 m/s in order to clear a 10 m wall that is 100 m away. Using the equations vf=vo+at, x=vot+1/2at^2, and vf^2=Vo^2+2ax, two equations can be derived to find the launch angle (θ) and time (t). By eliminating t, a quadratic equation for θ is obtained, with two possible solutions.
  • #1
trigunesq
1
0

Homework Statement


A projectile is launched at 36 m/s and travels 100 m horizontally. If it is to be able to just barely clear a 10 m wall, between what two angles should the projectile be launched with respect to the horizontal?




Homework Equations


vf=vo+at
x=vot+1/2at^2
vf^2=Vo^2+2ax
 
Physics news on Phys.org
  • #2
Welcome to PF!

Hi trigunesq! Welcome to PF! :smile:

(have a theta: θ and try using the X2 and X2 tags just above the Reply box :wink:)

They're asking you for the two angles from which it will be 10m up when it's 100m along.

Call the launch angle θ, and the time taken t, then use your constant acceleration equations for the x and y directions separately (a = 0 for the direction, of course), to get a pair of equations from which you can eliminate t to get a quadratic equation for θ (with two solutions! :wink:).

What do you get? :smile:
 
  • #3

θ=arctan(vy/vx)



The launch angle of a projectile is determined by the initial velocity and the height and distance it needs to travel. In this case, the initial velocity is 36 m/s and the horizontal distance is 100 m. We can use the equations vf=vo+at and x=vot+1/2at^2 to solve for the time it takes for the projectile to reach the wall.

First, we need to determine the vertical component of the initial velocity, which we can find using the equation vf^2=Vo^2+2ax. Since the projectile needs to clear a 10 m wall, the final vertical velocity will be 0 m/s. Solving for Vo, we get Vo=√(vf^2-2ax)=√(0^2-2(-9.8)(10))=14 m/s.

Next, we can use the equation vf=vo+at to solve for the time it takes for the projectile to reach the wall. Since vf=0 m/s and vo=14 m/s, we get 0=14+(-9.8)t, which gives us t=1.43 seconds.

Now, we can use the equation x=vot+1/2at^2 to solve for the horizontal distance traveled. Plugging in the values, we get 100=(14)(1.43)+1/2(-9.8)(1.43)^2. Solving for the initial launch angle, we get θ=arctan(vy/vx)=arctan(14/36)=21.8 degrees.

Therefore, the projectile should be launched at an angle between 21.8 degrees and 90 degrees with respect to the horizontal in order to just barely clear the 10 m wall. It is important to note that this solution assumes no air resistance and a flat surface. Real-world factors may affect the actual launch angle needed.
 

FAQ: Angular Velocity: Find Launch Angles for Projectile

What is angular velocity?

Angular velocity is the rate at which an object rotates or moves around a fixed point, measured in radians per second. It is a vector quantity, meaning it has both magnitude and direction.

How is angular velocity calculated?

Angular velocity is calculated by dividing the change in angle (in radians) by the change in time. The formula is: ω = Δθ/Δt, where ω is the angular velocity, Δθ is the change in angle, and Δt is the change in time.

How is angular velocity related to projectile motion?

In projectile motion, the object's path is a parabola and the object experiences both linear and angular motion. Angular velocity is related to projectile motion because it affects the object's trajectory and determines the launch angle required for the object to reach a specific distance.

How do you find launch angles for projectile using angular velocity?

To find the launch angle for a projectile, you can use the equation θ = tan^-1(vy/vx), where θ is the launch angle, vy is the vertical component of the projectile's velocity, and vx is the horizontal component of the projectile's velocity. You can also use trigonometry to solve for the launch angle.

How does air resistance affect the launch angle for projectiles?

Air resistance affects the launch angle for projectiles because it decreases the projectile's velocity and alters its trajectory. To compensate for this, a higher launch angle is required to achieve the desired distance. In general, the steeper the launch angle, the less the effect of air resistance on the projectile's motion.

Back
Top