# Initial speed with Angle and distance

1. Mar 11, 2013

### Computerguy

I have been working this one all day. No luck as far.

The question:

A projectile is launched at an angle of 53.1° with respect to the horizontal, as shown in Figure 2-34. ( Ill try to give you a picture of example 2-34) It strikes a target that is 120m away horizontally and 160m below the launching point. Find the initial speed of the projectile. Assume g=10m/s^2.

the photo should be in an attachment somewhere.

I would normally split this into several questions. But sadly that didn't help me much.

As far as i know i have theta and the distance in the X dirextion.

I have tried all my kinematic equations but they all want to have Time.

Accept for V^2=Vo^2+2a(X-Xo)

I try to use it. V=0 Vo= Unkown 2a(X-Xo) cancel due to no gravity in the x direction.

So basicly 0=Vo^2 hahaha

From research on the internet i heard there was a Range Formula.

x=Vo^2(sin2θ)/g

Sadly that only applies for flat surfaces. Also Why would g be there? There is no g in the x direction as far as i can tell.

At last i found the full equation.

x = Vo cosθ (Vo sinθ + √Vo^2 sin^2θ + 2g Yo )/g

Any suggestions? How do i find Distance in the y direction?

Thanks

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2. Mar 11, 2013

### Von Neumann

In the x-direction there is constant velocity, and in the y-direction there is constant acceleration. So, in the x-direction it's okay to apply constant velocity formulas, while in the y-direction it's okay to apply constant acceleration formulas.

3. Mar 11, 2013

### voko

Assume for a second that you are holding a ball just above a cliff that is 160 meters deep. You throw the ball vertically with some velocity v. How long does it take the ball to touch the bottom of the cliff?

4. Mar 11, 2013

### Von Neumann

Also for clarity, gravity is never not acting on the launched projectile; it is always present in this case.

5. Mar 11, 2013

### Computerguy

Thanks for the reply voko.

A think i have tried that before.

The canon shots the ball up from 160 m when it come back down to 160m it is the same V as when it came out of the canon.

Witch is the answer they are looking for.

If i knew the V i could solve it. And if i knew any info about time i could solve it.

I dont know how i would solve (Throwing it down) if i dont have a Vo a time or a distance in X direction. now the only variables i have is G and distance in y direction.

6. Mar 11, 2013

### Computerguy

Von neumann In the Y direction yes But g never effects anything in the x direction.

If you were to look at the x direction only you would see that v is constant. Meaning no g.

I am attempting to split the launched projectile in the x and y direction

7. Mar 11, 2013

### voko

That is correct, but I am asking for the time it takes to go from zero meters to some z meters up, and then to -160 meters down. Assume some initial velocity v.

If you still can't solve this, think how long it will travel till the highest point. What's so special about the highest point?

8. Mar 11, 2013

### Computerguy

I have though about it some more.

when it gets launched up Z meters it hits its highest point. At the highest point it is going at a V of 0m/s

From there it begins to drop a total distance of 160m + Z

Falling 160 m it travels for 5.7 s at a V 57m/s If i could only figure out distance Z i could solve the whole thing.

Witch kind of sends me back to the beginning. Z or Total Time are the key things i need to solve this. If i knew ether of these it would be solved.

I am stumped.

P.s. Right in my text book next to the question is the answer. It is 25m/s What i am more concerned of is how did he get to that answer.

Did you guys get that answer too?

Maybe i need to re read something. I must be missing a valuable piece of info.

How do i use the 53.1 degree angle they gave me? It must mean something. I dont have any kinematic equations that would let me use theta

9. Mar 11, 2013

### voko

At the highest point the vertical velocity is indeed zero. Now, assuming you know the initial vertical velocity is some v, and the acceleration due to gravity is g, how long does it take to reach the highest point?

10. Mar 11, 2013

### Computerguy

Quote: (Now, assuming you know the initial vertical velocity is some v)

Initial velocity in the y = Vo sin 53.1°

Still 2 unknowns.

I am missing something.

Please be patient with me. Im new to physics and dont understand everything yet.

11. Mar 11, 2013

### voko

Assume that v = Vo sin 53.1 is KNOWN. Find out the time to the highest point symbolically.

12. Mar 11, 2013

### Von Neumann

"Sadly that only applies for flat surfaces. Also Why would g be there? There is no g in the x direction as far as i can tell." ~ is what you said. I am simply trying to aware you of something that will confuse/anger physicists. Yes, the affect of g is only the vertical direction. It is correct to think of it component-wise and split the problem into two manageable parts.

As voko recommends, either solve for the time t in terms of other known terms and substitue, or just calculate it separately.

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