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A Projectile Fired At A 45 Degree Angleby miniradman
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#1
Oct411, 05:24 AM

P: 186

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. 


#2
Oct411, 05:52 AM

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#3
Oct411, 05:52 AM

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step 1. Write down all of the equations you know that might be useful here.



#4
Oct411, 06:02 AM

P: 186

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 


#5
Oct411, 06:15 AM

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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.



#6
Oct411, 06:17 AM

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#7
Oct411, 06:23 AM

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


#8
Oct411, 06:34 AM

P: 186

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 


#9
Oct411, 06:42 AM

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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. 


#10
Oct411, 06:47 AM

P: 186

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? 


#11
Oct411, 07:04 AM

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? :) 


#12
Oct411, 07:09 AM

P: 186

I think so, but I don't see how you can get numbers out of those letters



#13
Oct411, 07:12 AM

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. 


#14
Oct411, 08:10 AM

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. 


#15
Oct411, 08:51 AM

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#16
Oct411, 08:55 AM

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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 :)



#17
Oct411, 04:05 PM

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#18
Oct411, 07:17 PM

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