Projectile Motion find the highest angle

In summary, the conversation discusses finding the highest angle at which a projectile will clear a peak. Various equations are presented to solve for the time and distance at the top of the parabola, but it is important to note that the top of the trajectory may occur before, at, or after the peak is encountered as long as the projectile's altitude clears the peak.
  • #1
kukumaluboy
61
1

Homework Statement



http://postimg.org/image/7yfw1zpq1/

http://postimg.org/image/7yfw1zpq1/ <-Question

The Attempt at a Solution


I don't really understand the question but ill try.

I need to first find the highest angle at which the projectile will clear the peak.
At Peak:

Sy = 0.5(u+Vpeak)t , vpeak = 0
1800 = 125tsinx
tsinx = 14.4
t = 14.4/sinx -equation1

Sx = 0.5+(Ux+Vpeakx)t , Ux=Vx i guess?
2500 = 0.5(250cosx)t
2500 = 125tcosx
tcosx = 20
t = 20/cosx
Substitute eq 2 into eq 1:
20/cosx = 14.4/sinx
cosx = 1.39sinx

Then i don't know how to solve
 
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  • #2
You need to rethink a bit. It is not necessary that the vertical velocity is zero when passing the peak. Try finding the height as a function of the horizontal distance for an arbitrary angle.
 
  • #3
need more tips lol
 
  • #4
This is a nice problem. I agree with Orodruin, that is probably the easiest method. What is your difficulty in following it?
 
  • #5
At top of parabola Vy = 0

Vy = Uy + at
0 = 250sinx + 9.81t
t = 25.5sinx

After that?
 
  • #6
kukumaluboy said:
At top of parabola Vy = 0

Yes, but the top of the parabola is not necessarily where the projectile passes the mountain top.
 
  • #7
Orodruin said:
Yes, but the top of the parabola is not necessarily where the projectile passes the mountain top.

The top of the trajectory of the projectile can be reached before the peak is encountered, at the peak, or after the peak is encountered, so long as the altitude of the projectile clears the top of the peak.
 
  • #8
SteamKing said:
The top of the trajectory of the projectile can be reached before the peak is encountered, at the peak, or after the peak is encountered, so long as the altitude of the projectile clears the top of the peak.

How does this contradict what I just said?
 

FAQ: Projectile Motion find the highest angle

1. What is projectile motion?

Projectile motion is the motion of an object through the air that is affected by gravity. It is a combination of horizontal and vertical motion, where the horizontal motion is constant and the vertical motion is affected by the force of gravity.

2. How do you find the highest angle in projectile motion?

The highest angle in projectile motion is known as the maximum height, which occurs when the vertical velocity is zero. To find this angle, you can use the equation v = v0sinθ - gt, where v0 is the initial velocity, θ is the angle, and g is the acceleration due to gravity. Set the equation equal to zero and solve for θ to find the highest angle.

3. What factors affect the highest angle in projectile motion?

The highest angle in projectile motion is affected by the initial velocity, the angle of projection, and the acceleration due to gravity. A higher initial velocity or a lower angle of projection will result in a higher maximum height. Additionally, a higher acceleration due to gravity will result in a lower maximum height.

4. How does air resistance impact projectile motion?

Air resistance, also known as drag, can impact projectile motion by slowing down the object and changing its trajectory. This effect is more significant for objects with a larger surface area and traveling at higher velocities. In most cases, air resistance is negligible and can be ignored when calculating projectile motion.

5. How is projectile motion used in real life?

Projectile motion is used in a variety of real-life scenarios, such as throwing a ball, shooting a basketball, or launching a rocket. It is also used in sports and activities like baseball, golf, and archery. Understanding projectile motion can also help engineers design and launch projectiles, such as missiles and satellites.

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