Projectile Motion Problem

1. Jan 30, 2013

RjD12

1. The problem statement, all variables and given/known data

At a time t, a projectile is measured to have a height Y = 12 m above the ground and a velocity v = 4 m/s at an angle (theta) = 42 with respect to to the horizontal. At what horizontal positions is the particle at a height 3/4 Y ?

Height= 12 m
Velocity= 4 m/s
Angle= 42 degrees

2. Relevant equations

I believe both the range equation x= x0 + V0xt and y= y0 + V0yt - (1/2)gt2

3. The attempt at a solution

I tried to to use the second equation, but I had trouble with time since it was not given. I used 4sin(42) for V0y, but I don't think it helped me much.

This should be simple, I can tell, but I am just stuck.

2. Jan 30, 2013

MechTech

You are on the right track with your attempt but you need to find a relationship between the two equations. (HINT: What do the x position and Y position equations have in common)

3. Jan 30, 2013

RjD12

They both have initial velocities, and those can be figured out using the velocity given and the angle. X0 would be 4cos(42) and Y0 would be 4sin(42).

Am I getting any closer?

4. Jan 30, 2013

MechTech

Think variable. Both functions are dependant on time.

5. Jan 30, 2013

RjD12

Would I have to set them equal to each other by solving for time? I know that the time is the same for both directions.

6. Jan 30, 2013

RjD12

I also recently tried using the quadratic formula with my Y equation, but I end up getting a negative under the radical. I am not sure why.

7. Jan 30, 2013

MechTech

Remember that the generic quadratic eq is ax^2+bx+c. You should get a positive number

8. Jan 30, 2013

RjD12

If I use y= y0 + V0yt - (0.5)gt2

I get

12= 0 + 4sin(42)t-(9.8/2)t2

In the general equation form, it is:

-4.9t2+4sin(42)-12

9. Jan 30, 2013

MechTech

y_0 is 12 not zero. Y should be 3/4

10. Jan 30, 2013

RjD12

So what would be the final Y position? It seems like since it is in motion, it would start at zero.

11. Jan 30, 2013

MechTech

From the problem we can assume that at t=0 the Y position is 12m. The last line ask us to solve from is 3/4 of Y or 0.75*12. Thus your equation for Y should look as follows:
0.75 *12 = 12 + 4* sin(42)*t - 9.8/2*t^2

12. Jan 30, 2013

RjD12

Okay. Yes, that looks like what I have after fixing it. I must be creating some sort of error in my calculations then, because I plugged the equation : -4.9t2 + 4sin(42)*t +3 into a graphing calculator, and there is indeed a positive value.

It is odd. Using the quadratic formula, underneath the radical, which is sqrt(b^2-4ac),
b^2= (4sin(42))^2, but 4ac is a bigger number, being : 4 (-4.9)(3)= 58.8.

I might just be doing something wrong I suppose.

13. Jan 31, 2013

MechTech

You are forgetting you negative sign. A = -9.8