# How to find initial velocity from range and angle.

1. Mar 14, 2009

### iHEARTmath

1. The problem statement, all variables and given/known data
A golf ball is hit at an angle of 30 degrees, it travels 50m landing straight in the hole. Ignoring air friction, what is the initial velocity?

2. Relevant equations
not too sure, d=vi.t+1/2.a.ts∧2, vf=u∧2+2.a.d, v=u+a.t, d=u.t

3. The attempt at a solution
Vertical initial Velocity= usin30
Horizontal initial velocity= ucos30

2. Mar 14, 2009

### LowlyPion

Welcome to PF.

That's a good start. Maybe try to eliminate time from your equations?

3. Mar 14, 2009

### iHEARTmath

Could you give some advice in how I would do that? Like which formula to use?
Can it be figured out using these formulas?

4. Mar 14, 2009

### iHEARTmath

t= d/u, so t=d/ucos30. Do I leave it as that or.. t= 50/ucos30?

5. Mar 15, 2009

### The Liberator

remember that when there is no air resistance the vertical and horizontal components of the velocity are completely separate from each other, and because there is no air resistance, the horizontal velocity stays the same throughout the flight.

P.S. Have I said too much?*

*this is to the admins and mentors and other senior members

6. Mar 15, 2009

### iHEARTmath

I tried working it out...
Uv= usin30
Uh= ucos30
t= d/Uh, t= 50/ucos30 s=Uv.t+(1/2).-9.8∧2
and does s=0? I can't remember why.
0=usin30.(50/ucos30)+(1/2).-9.8.(50/ucos30)∧2
0=??? how do i simplify this?

7. Mar 15, 2009

### The Liberator

maybe try and refer to earlier notes on this, that the book might have talked about, to do with components of vertical/horizontal linear motion…

8. Mar 15, 2009

### iHEARTmath

if s=0 it's simplified to u= (square root of 9.8x50)/cos30 and the answer is 23.78661943
but if s=50 the answer is 27.80110847
What is correct? and why?

9. Mar 15, 2009

### The Liberator

okay then.

When s=25, then v will be at it's minimum value, and there will be no vertical velocity at all. you can then work out what the vertical distance of s is (with the help of tan 30). from that answer, you can then use t = sv…

…then everything else will slowly fall into place.

10. Mar 15, 2009

### iHEARTmath

Ah.. what do I use the vertical height for? 25tan30= 14.43375673
so then,
Uv=0
Height?=14.43375673

11. Mar 15, 2009

### Phrak

Good grief, golf balls are not pojectiles. Fire your professor.

12. Mar 15, 2009

### iHEARTmath

Vf=Vi+-9.8x(25/Vi.cos30)
0=v-9.8x(25/v.cos30)
v=9.8x(25/v.cos30)
u∧2cos30=9.8x25
u∧2=(9.8x25)/cos30
u= square root((9.8x25)/cos30)
u=16.8196799??

13. Mar 15, 2009

### The Liberator

the vertical height can be used to find the flight time for the golf ball half way through the overall flight.

With that you can then find the initial velocity of the ball.

14. Mar 15, 2009

### iHEARTmath

What is the formula that is needed to find the flight time for half of the journey?

15. Mar 15, 2009

### The Liberator

If you have the vertical displacement, then it would be this: t=sv

As s = 25 m, and at the halfway point of the flight time, the golf ball is at it's highest point, the vertical velocity is 0 ms-1, then you can do the maths.

Also, as a reference, here are the formulae.

16. Mar 15, 2009

### iHEARTmath

Sorry I don't understand, t=sv; t=25x0? t=0?

17. Mar 15, 2009

### The Liberator

No, no, it is me. There are other equations:

$$s=\frac{1}{2}(u+v)t$$
$$v=u+at$$
$$s=ut+\frac{1}{2}at^2$$
$$v^2=u^2+2as$$

Which one is the most appropriate? Also, "v" is the final and "u" is the initial velocity. from knowing the answer to the equation you use, you then just need to do a bit of trig…

18. Mar 15, 2009

### LowlyPion

It's important to keep the vertical and horizontal components separate. In the horizontal direction happily you are given the distance X as 50m.

The equation "s=Uv.t+(1/2).-9.8∧2" is more clearly written as

Y = VoSinθ*t - 1/2*g*t2

S or Y as I have rewritten it is 0 because you end at net 0 Y distance up/down. It is important not to confuse X and Y.

So in Y, VoSinθ*t = 1/2*g*t2 and this yields pretty simply that

t = 2VoSinθ/g

But you also know from the horizontal X direction that t = X/VoCosθ

Now you should find yourself in splendid position to eliminate t as suggested, since you happily have two equations, albeit 1 horizontal and the other vertical, that must have t = t.

You don't need to be Tiger Woods to get to the 19th tee now I should hope.

19. Mar 15, 2009

Thank you:)