A Projectile Fired At A 45 Degree Angle

Click For Summary
SUMMARY

The discussion focuses on calculating the initial velocity of a projectile fired at a 45-degree angle that just clears a 6-meter high fence located 100 meters away. Participants emphasize the importance of breaking down the problem into horizontal and vertical components using kinematic equations. Key equations discussed include the horizontal component \( V_{0x} = V_0 \cos(45^\circ) \) and the vertical component \( V_{0y} = V_0 \sin(45^\circ) \). The solution involves simultaneous equations derived from the projectile motion equations, leading to the determination of the initial velocity \( V_0 \).

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions (sine and cosine)
  • Ability to solve simultaneous equations
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to apply trigonometric functions to resolve vector components
  • Practice solving simultaneous equations in physics problems
  • Explore the effects of varying launch angles on projectile trajectories
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to enhance their teaching methods in these topics.

  • #31
6=9.8t^2*(100/0.7071t)*0.7071

we just told you that you should have a plus in there, not a multiplication!

you do something wrong when you insert the substitution.
 
Physics news on Phys.org
  • #32
Ah, found the problem,

s = ut+(1/2) at^2

I used this formula and I got the right answer... thanks anyway guys
 
  • #33
well you can't solve this problem with only one equation.
 
  • #34
Yep, I used that s=ut+1/2 at^2 for both the horizontal and vertical components ;)
 
  • #35
With 0 acceleration in the horionztal I take it. That's actually what I said in my equations.
 
  • #36
Yes true, but there was still no 1/2 in either of them.

I tried to work it out without the 1/2 and I kept getting a cubic which cannot be solved. The one half made heaps of difference ;)
 
  • #37
OUCH! i simply forgot to write down the half! Pretty amazing that no-one noticed! Sorry about that, dude! :)(the half actually appears from taking the anti-derivative of of t, resulting in 0.5(t)^2.)
 
Last edited:
  • #38
I've haven't learned about anti-derivatives yet. But I see what you have done there.

Its not a problem mate, everyone makes mistakes ;)
 
  • #39
Yup. Those 4 basic equations of motions is what you use when you start out with physics, but you see clearly where they naturally come from first when you learn about derivates and integration.

:)
 

Similar threads

Find Projectile Flight Time Given Only Maximum Height
  • · Replies 4 ·
Replies
4
Views
3K
Optimal Angle and Horizontal Range of a Bottle Rocket Projectile
  • · Replies 1 ·
Replies
1
Views
3K
How Long to Reach Maximum Height for a 70 m/s Projectile at 45 Degrees?
  • · Replies 5 ·
Replies
5
Views
2K
A plane, diving with constant speed at an angle of 53.0 degrees....
  • · Replies 2 ·
Replies
2
Views
3K
Projectile motion of a cannon ball versus time
  • · Replies 15 ·
Replies
15
Views
2K
How Do You Solve a Projectile Motion Problem?
  • · Replies 3 ·
Replies
3
Views
4K
Projectile Motion on the Moon: Finding the Optimal Launch Angle Using Equations
  • · Replies 2 ·
Replies
2
Views
3K
Initial Position Discrepancy Involving Fluid Resistance
  • · Replies 5 ·
Replies
5
Views
2K
Projectile Motion - object that is shot at an angle of 55 degrees
  • · Replies 8 ·
Replies
8
Views
2K
Stone being tossed from mountain Uni Physics with Calc
  • · Replies 14 ·
Replies
14
Views
2K