A Projectile Fired At A 45 Degree Angle

192
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. Hootenanny

9,681
Staff Emeritus
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.

Staff: Mentor

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

192
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?

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

Last edited: Oct 4, 2011

Staff: Mentor

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.

192
Sorry mate, but I'm not exactly sure what is meant by "in terms of"

Staff: Mentor

Okay.

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

Now write the vertical component.

192
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

Staff: Mentor

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.

192
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. Cipherflak

33
if the initial speed is v0, the horizontal component is u0=v0*cosθ and the vertical w0=v0*sinθ.

Do you see this? :)

Last edited: Oct 4, 2011

192
I think so, but I don't see how you can get numbers out of those letters

13. Cipherflak

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. Cipherflak

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.

Last edited: Oct 4, 2011

Staff: Mentor

Yes, it does look like 3 unknowns, but since we were told that theta is 45 degrees, then that leaves only two unknowns.

16. Cipherflak

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 :)

192
So simultaneous equations?

Yes.