Projectile Motion Problem: Finding Horizontal Position at a Given Height

[RjD12]

Homework Statement

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

Homework Equations

I believe both the range equation x= x₀ + V_0xt and y= y₀ + V_0yt - (1/2)gt²

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 V_0y, but I don't think it helped me much.

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

 

[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)

 

[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?

 

[MechTech]

Think variable. Both functions are dependant on time.

 

[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.

 

[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.

 

[MechTech]

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

 

[RjD12]

If I use y= y₀ + V_0yt - (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

Unless this is incorrect, my answer is negative under the radical.

 

[MechTech]

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

 

[RjD12]

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

 

[MechTech]

Homework Statement

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

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

 

[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.

 

[MechTech]

You are forgetting you negative sign. A = -9.8