# A Projectile Fired At A 45 Degree Angle

Tags: projectile
 P: 181 1. The problem statement, all variables and given/known data A projectile is fired at a 45 degree angle and its just barely able to make it above a 6m high fence that is 100 meters away. What was the projectile's initial velocity? 2. Relevant equations x and y components? 3. The attempt at a solution Well the problem with this one is that I don't know where to start. I mean this is basically the reverse of everything I've learnt about projectile motion. I think I should be calculating the x and y components first but I'm not even sure how to do that.
Emeritus
PF Gold
P: 9,772
 Quote by miniradman 1. The problem statement, all variables and given/known data A projectile is fired at a 45 degree angle and its just barely able to make it above a 6m high fence that is 100 meters away. What was the projectile's initial velocity? 2. Relevant equations x and y components? 3. The attempt at a solution Well the problem with this one is that I don't know where to start. I mean this is basically the reverse of everything I've learnt about projectile motion. I think I should be calculating the x and y components first but I'm not even sure how to do that.
In all these constant acceleration kinematics problems, the best place to start is by writing two lists. The first is a list of everything that you know already. The second, is a list of the things you want to know. You should do this for each component (vertical and horizontal) of the motion.
 HW Helper Thanks P: 5,246 step 1. Write down all of the equations you know that might be useful here.
 P: 181 A Projectile Fired At A 45 Degree Angle 1. v=u+at (cannot use because there is no time or final velocity) 2. s= (u+v)/2 x t (cannot use because no time, distance or inital and final velocity) 3. s= ut + 1/2 at^2 (cannot use because there is no initial or final velocities or time or distance) 4. v^2 = u^2 +2as (cannot use this one because we have no inital or terminal velocity or distance) Others s=d/t (I'll most likely (definately) use this at the end when I have my horizontal and veritcle components) I'm sorry, but I honestly dont see how I can just have 1 variable when using any of these kinematics (when solving for the horizontal and veritcle components). Could using simultaneous equations be answer? What I already know: Distance from fence to launch point = 100m Gravitational Acceleration = -9.8m/s or 9.8m/s Angle of launch = 45 degrees the x and y components would be equal? What I need to know: x and y components total distace covered while in the air time of flight? inital velocity
 HW Helper Thanks P: 5,246 It's fired with a velocity V at 45 degrees. Write the horizontal component of firing speed, in terms of V. Write the vertical component of firing speed, in terms of V.
P: 181
 Quote by NascentOxygen It's fired with a velocity V at 45 degrees. Write the horizontal firing speed, in terms of V. Write the vertical firing speed, in terms of V.
Sorry mate, but I'm not exactly sure what is meant by "in terms of"
HW Helper
Thanks
P: 5,246
 Quote by miniradman Sorry mate, but I'm not exactly sure what is meant by "in terms of"
Okay.

It's fired with a velocity V at 45 degrees. Write the horizontal component of that.

Now write the vertical component.
 P: 181 So the verticle component would be Sinθ = o/V Sin(45 = o/V The horizontal component Cosθ = a/V Cos(45 = a/V where hypotenues is equal to V what now? this is where I get stuck because I don't know if I have enough infomation to move on or am I missing something
 HW Helper Thanks P: 5,246 What is o? What is a? This topic reserves a for acceleration. If you use the same symbols for different things, you will soon get confused. Besides, I can't see your expression for the horizontal velocity.
 P: 181 sorry, I was using trig functions where o = opposite and a = adjacent. I might just keep it as: a = adj and o = opp Can I use Cosθ = adj/hyp or speed=distance/time to figure out the horizontal component?
 P: 33 if the initial speed is v0, the horizontal component is u0=v0*cosθ and the vertical w0=v0*sinθ. Do you see this? :)
 P: 181 I think so, but I don't see how you can get numbers out of those letters
 P: 33 one step at a time ;) you may be used to other notations like v_x instead of u or whatever, but get comfortable with what components mean first.
 P: 33 this excerise is a bit tricky, cause as you say, it's "the other way around". but we know, considering the equations for the two components, that; 6m= - gt^2+V0*sinθ*t (the relevant velocity is here the vertical hence the sine) 100m= V0*cosθ*t (the horizontal velocity calls for the use of a cosine) everybody concur? :) Since we have 2 equations and 2 unknowns, finding the answer for V0, the initial velocity, is just some algebraic puzzlework.
HW Helper
Thanks
P: 5,246
 Since we have 2 equations and 2 unknowns, finding the answer for V0, the initial velocity, is just some algebraic puzzlework.
Yes, it does look like 3 unknowns, but since we were told that theta is 45 degrees, then that leaves only two unknowns.
 P: 33 You're absolutely right, I forgot to insert for theta. It's best to keep the symbols as long as possible though, to obtain a more general solution :)
P: 181
 Quote by Cipherflak this excerise is a bit tricky, cause as you say, it's "the other way around". but we know, considering the equations for the two components, that; 6m= - gt^2+V0*sinθ*t (the relevant velocity is here the vertical hence the sine) 100m= V0*cosθ*t (the horizontal velocity calls for the use of a cosine) everybody concur? :) Since we have 2 equations and 2 unknowns, finding the answer for V0, the initial velocity, is just some algebraic puzzlework.
So simultaneous equations?
HW Helper
Thanks
P: 5,246
 Quote by miniradman So simultaneous equations?
Yes.

 Related Discussions Introductory Physics Homework 13 Introductory Physics Homework 3 Introductory Physics Homework 7 Introductory Physics Homework 2 Introductory Physics Homework 3