# Projectile Motion calculation problem

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1. Feb 20, 2016

### x2017

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

2. Relevant equations
D=VΔt
Vf=Vi+ayΔt
Df=Di+VyΔt+(1/2)ayΔt2

3. The attempt at a solution
I solved for d using 1/2ayΔt2 and got 10.30m (Solving for d is a question later on in this assignment and i checked it there and it is incorrect...)

Then I used
Df=Di+VyΔt+(1/2)ayΔt2
but every time I try this equation I get the wrong answer because I am unable to calculate d correctly...

I'm not sure if I'm stuck because I can't get d right and need d for Vxinitial or if I'm just going about solving for Vxinitial all wrong.

2. Feb 20, 2016

### cnh1995

Look at the diagram carefully. What parameters of the projectile do you know from the diagram?

3. Feb 20, 2016

### J Hann

Try breaking the problem into solvable segments.
What vertical velocity is required to reach a height of 2 h?
How long does it take to reach that height?
What horizontal velocity is required, etc.?

4. Feb 20, 2016

### x2017

I know the lengths of x, h, 2h, could get the hypotenuse of the x/2h triangle...

I have:
Vyinitial=12.05m/s
Time for the ball to reach max height above the tree=1.23s
Time for ball to travel from max height above tree to the surface of the upper green=0.87s
from previous questions. Meant to put this info up top in the first post (all three of these numbers are correct).

5. Feb 20, 2016

### cnh1995

Are you sure? I'm getting a different answer. You have maximum height h of the projectile as 3.7m.

6. Feb 20, 2016

### x2017

Yep!

Dyfinal=Dyinitial+VyinitialΔt+(1/2)ayΔt2
3.7=0+VyinitialΔt+(1/2)(9.81)Δt2

Δt=[(2Δdy)/g]
Δt=[(2)(3.7)/9.81]1/2
Δt=1.23

7.4=0+Vyinitial(1.23)+(1/2)(9.81)(1.23)2
14.82/1.23=Vyinitial
12.05m/s=Vyinitial

Max height is 2h isn't it? So it'd be 7.4m

7. Feb 20, 2016

### cnh1995

Oh sorry! I didn't read the h in the green part of the tree.. 12.05m/s it is.. Did you get Vx?

8. Feb 20, 2016

### x2017

No worries! :)
No, I'm stuck on Vxinitial, I got the y-component right away so I'm confused as so why I am unable to get the x-component!
I using this equation Dyfinal=Dyinitial+VyinitialΔt+(1/2)ayΔt2 again but nothing I try works out...

9. Feb 20, 2016

### cnh1995

You have Vyinitial=12.05m/s. You can calculate the angle of projection θ using the height and x. Since Vyinitial=Vinitialsinθ, you can get Vinitial.

10. Feb 20, 2016

### cnh1995

Not by pythagoras and trigonometry. Here, x is half of the range(R/2)of the projectile and 2h is the maximum height. There is a relation between R, H and θ. You can get θ from it and proceed.

11. Feb 20, 2016

### x2017

Phew, okay thanks! I was trying with pythagoras/trig and was like "why am I not getting it?????"
I'll try this way now!

12. Feb 20, 2016

### cnh1995

There's one simpler way. You can know the time taken to reach the maximum height. Using that time and horizontal distance x, you can calculate Vx. Vx is constant throughout. The relation between R,H and θ is Rtanθ=4H.

Last edited: Feb 20, 2016
13. Feb 20, 2016

### x2017

I wasn't getting it so I just tried Vx=dx/t as a last ditch effort and that worked......
Vx=10.4/1.23
Vx=8.46m/s

However, for the next part of the question
"A golfer must hit a pitch shot over a tree and onto an elevated green. Assuming air resistance is negligible, what initial launch angle must be imparted to the ball so that it will follow the trajectory indicated in the figure above? Answer in degrees above the horizontal."
Using the inverse tan doesn't work so would I use the method you're talking about?

EDIT:
never mind, it did work with the trig. I had to use the velocities not the distances (duh)
Thanks for all of your help!!

14. Feb 20, 2016

### cnh1995

You're welcome!
θ=tan-1(Vy/Vx)..
The relation between R, H and θ i.e. Rtanθ=4H can also be used as one of the standard formulae..