Projectile Motion calculation problem

[x2017]

Homework Statement

Homework Equations

D=VΔt
V_f=V_i+a_yΔt
D_f=D_i+V_yΔt+(1/2)a_yΔt²

The Attempt at a Solution

I solved for d using 1/2a_yΔt² 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
D_f=D_i+V_yΔt+(1/2)a_yΔt²
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 V_xinitial or if I'm just going about solving for V_xinitial all wrong.

 

[cnh1995]

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

 

[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.?

 

[x2017]

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

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

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.?

I have:
V_yinitial=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).

 

[cnh1995]

Vyinitial=12.05m/s

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

 

[x2017]

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

Yep!

D_yfinal=D_yinitial+V_yinitialΔt+(1/2)a_yΔt²
3.7=0+V_yinitialΔt+(1/2)(9.81)Δt²

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

7.4=0+V_yinitial(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

 

[cnh1995]

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

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

 

[x2017]

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

No worries! :)
No, I'm stuck on V_xinitial, 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 D_yfinal=D_yinitial+V_yinitialΔt+(1/2)a_yΔt² again but nothing I try works out...

 

[cnh1995]

You have V_yinitial=12.05m/s. You can calculate the angle of projection θ using the height and x. Since V_yinitial=V_initialsinθ, you can get V_initial.

 

[cnh1995]

You can calculate the angle of projection θ using the height and x.

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.

 

[x2017]

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.

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

 

[cnh1995]

Phew, okay thanks! I was trying with pythagoras/trig and was like

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.

 

[x2017]

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 Rsinθ=4H.

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!

 

[cnh1995]

Thanks for all of your help!

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..