How to find initial velocity from range and angle.

In summary, when s=25, the vertical velocity is 0 ms-1 and there is no vertical displacement, so the flight time is t=sv.
  • #1
iHEARTmath
10
0

Homework Statement


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?


Homework 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


The Attempt at a Solution


Vertical initial Velocity= usin30
Horizontal initial velocity= ucos30
 
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  • #2
iHEARTmath said:

Homework Statement


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?

Homework 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

The Attempt at a Solution


Vertical initial Velocity= usin30
Horizontal initial velocity= ucos30

Welcome to PF.

That's a good start. Maybe try to eliminate time from your equations?
 
  • #3
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
t= d/u, so t=d/ucos30. Do I leave it as that or.. t= 50/ucos30?
 
  • #5
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
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
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
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?
Please help
 
  • #9
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
Ah.. what do I use the vertical height for? 25tan30= 14.43375673
so then,
Uv=0
Height?=14.43375673
 
  • #11
Good grief, golf balls are not pojectiles. Fire your professor.
 
  • #12
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
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
What is the formula that is needed to find the flight time for half of the journey?
 
  • #15
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
Sorry I don't understand, t=sv; t=25x0? t=0?
 
  • #17
iHEARTmath said:
Sorry I don't understand, t=sv; t=25x0? t=0?

:blushing:

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

[tex]s=\frac{1}{2}(u+v)t[/tex]
[tex]v=u+at[/tex]
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]v^2=u^2+2as[/tex]

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…

…and there is your answer.
 
  • #18
iHEARTmath said:
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.

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
Thank you:)
 

1. How do I calculate initial velocity from range and angle?

To find the initial velocity from range and angle, you can use the equation v0 = √(Rg)/sin(2θ), where v0 is the initial velocity, R is the range, g is the acceleration due to gravity, and θ is the launch angle.

2. What if I don't know the range or angle?

If you do not know the range or angle, you will not be able to calculate the initial velocity. In this case, you will need to gather more information or use other equations to solve for the initial velocity.

3. How does air resistance affect the initial velocity?

Air resistance can affect the initial velocity in various ways, depending on the shape and weight of the object being launched. In general, air resistance will decrease the initial velocity and therefore affect the range of the object. To account for air resistance, you can use more complex equations or conduct experiments to determine its impact.

4. Can I use the same equation for objects launched at different angles?

Yes, the equation v0 = √(Rg)/sin(2θ) can be used for objects launched at different angles. However, keep in mind that the launch angle will affect the range of the object, so you will need to adjust the equation accordingly.

5. What are some common units used for initial velocity?

The most common units for initial velocity are meters per second (m/s) or feet per second (ft/s). However, you can also use other units such as miles per hour (mph) or kilometers per hour (km/h), as long as the units are consistent with the other variables used in the equation.

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