Finding Initial Velocity for Projectile Launch at Angle Theta

[ChromoZoneX]

How to find the initial velocity required for a projectile to be launched at an angle theta, with a distance 'd' and height 'h'.

An expression is required. I am unable to find one using all three.

Thanks in advance.

 

[SammyS]

What do you mean by height? --- the maximum height achieved by the projectile?

In my opinion, it's generally best to solve such problems by considering that motions in the vertical & horizontal directions are independent, rather than looking for some "grand" solution for each possible scenario.

 

[ChromoZoneX]

Yeah, I mean maximum height achieved by the projectile. It's a spring that needs to be stretched 'x' so that it attains a certain initial velocity 'v' so that it hits a target 'd' meters away and 'h' meters high from the same level of launch.

I know that x = √(mv²/k)

I know the value of 'k'. I need an equation for the value of 'v'.

Any ideas?

 

[SammyS]

Let's see if I understand. The target is d meters away, horizontally, and is h meters higher in elevation than the launch position. You want to know what velocity, v, is needed to launch the projectile at an angle of θ and have the projectile hit the target. Is that correct?

 

[gneill]

 

[ChromoZoneX]

Yes that is correct. Given height, distance and angle of launch, find velocity.

PS: That is an awesome drawing!

 

[genericusrnme]

I'd start off by writing Newtons law for the projectile
F=ma

since the force is gravity we have F=mg and so for the y axis
a=g

you can either solve this differential equation or look up the solution, it should be in any intro physics textbook, and you'll find that the y position of the particle is;
y=a_0+a_1t+\frac{g}{2} t^2
and for the x position you'll find;
x=b_0+b_1t

you'd then set a_0 = b_0 = 0 to make the projectile be launched from the origin of co-ordinates
you set the velocity as V Sin and V Cos, which gives you;
y=V Cos( \theta ) t + \frac{g}{2} t^2
x=V Sin( \theta )t

so you now have to equations which you can solve, given some time you want to hit the target at, for V and theta
Tan( \theta )=\frac{x}{y-\frac{g}{2}t^2}
\theta = arctan(\frac{x}{y-\frac{g}{2}t^2})
where x and y are the x and y of the target

and you can substitute your theta back into find the V

if you already had the angle theta (I couldn't tell from your OP) you can put it in the tan equation and solve for t, then sub that into the x equation to get v

good luck!

 

[ChromoZoneX]

I have the angle theta, the distance the target is away and the height of the target. I don't have the time.

I need to find the initial velocity of the projectile.

From your equations,

I get the "v sin" and "v cos" equations for x and y.
Shouldn't the 'x' eqn use v cos(theta) and 'y' be v sin(theta)?

But where do I get the value of 't'?

Also, the tan(theta) doesn't give me the velocity?

 

[gneill]

I have the angle theta, the distance the target is away and the height of the target. I don't have the time.

I need to find the initial velocity of the projectile.

From your equations,

I get the "v sin" and "v cos" equations for x and y.
Shouldn't the 'x' eqn use v cos(theta) and 'y' be v sin(theta)?

Yes.

But where do I get the value of 't'?

Also, the tan(theta) doesn't give me the velocity?

Write the equations for the horizontal and vertical components of the motion. Use v cos(θ) for the initial horizontal velocity, and v sin(θ) for the initial vertical velocity. When the projectile is at the target, how will these equations look (what values are in place for the x and y variables)?

 

[ChromoZoneX]

OK this is my equation for 't',

t=\sqrt{}((xtan\vartheta - y)/(-0.5g))

Can I substitute this for 't' in the equation,

v = x/(cos\varthetat)

 

[gneill]

OK this is my equation for 't',

t=\sqrt{}((xtan\vartheta - y)/(-0.5g))

Can I substitute this for 't' in the equation,

v = x/(cos\varthetat)

Can you show how you arrived at your equation for t?

 

[ChromoZoneX]

x = vcos\vartheta.t

t = x/(vcos\vartheta)

tan\vartheta=(y-0.5gt²)/x

Isolating 't',

t² = (x tan\vartheta - y)/-0.5g

t = \sqrt{}((x tan\vartheta - y)/-0.5g)

 

[ChromoZoneX]

Also, when we talk about 'x', is it the total distance traveled by the projectile or is it just half the distance (that is the horizontal distance covered when it has reached its maximum height)?

 

[gneill]

x = vcos\vartheta.t

t = x/(vcos\vartheta)

tan\vartheta=(y-0.5gt²)/x

Where does this come from?

 

[gneill]

Also, when we talk about 'x', is it the total distance traveled by the projectile or is it just half the distance (that is the horizontal distance covered when it has reached its maximum height)?

It's whatever you want it to mean It's the value of x that corresponds to the corresponding value of y at the same instant of time.

So if the time of interest is the instant that the projectile is at the target, then x is the horizontal distance from the launch point to the target, and y is the vertical height of the target w.r.t. the launch point.

 

[ChromoZoneX]

I'd start off by writing Newtons law for the projectile
F=ma

since the force is gravity we have F=mg and so for the y axis
a=g

you can either solve this differential equation or look up the solution, it should be in any intro physics textbook, and you'll find that the y position of the particle is;
y=a_0+a_1t+\frac{g}{2} t^2
and for the x position you'll find;
x=b_0+b_1t

you'd then set a_0 = b_0 = 0 to make the projectile be launched from the origin of co-ordinates
you set the velocity as V Sin and V Cos, which gives you;
y=V Cos( \theta ) t + \frac{g}{2} t^2
x=V Sin( \theta )t

so you now have to equations which you can solve, given some time you want to hit the target at, for V and theta
Tan( \theta )=\frac{x}{y-\frac{g}{2}t^2}
\theta = arctan(\frac{x}{y-\frac{g}{2}t^2})
where x and y are the x and y of the target

and you can substitute your theta back into find the V

if you already had the angle theta (I couldn't tell from your OP) you can put it in the tan equation and solve for t, then sub that into the x equation to get v

good luck!

All I did is switch it to ,
y=V Sin( \theta ) t + \frac{g}{2} t^2
x=V Cos( \theta )t

and then divided y/x to get tanθ and hence the equation.

 

[gneill]

Okay, well this is really going the long way around!

y = v\;sin(\theta)\;t + \frac{1}{2} g t^2 (for some reason choosing a negative value for the constant g )
x = v\;cos(\theta)\;t

Then:
sin(\theta) = \frac{y - \frac{1}{2} g t^2}{v t} and

cos(\theta) = \frac{x}{v t}

tan(\theta) = \frac{y - \frac{1}{2} g t^2}{x}

Solving for t:

t = \sqrt{\frac{2}{g} (y - x\;tan(\theta)}

The reason I say that this is the long way around is that you can find a value for t very easily from the equation for the horizontal motion. Substitute that into the equation for the vertical motion and solve for v.

 

[ChromoZoneX]

Now, I'm confused. Let me explain the whole scenario.

I have to make a spring launcher that will fire a stretched spring. I will be given a target that is 'x' meters away and 'y' meters high w.r.t launch point. I will also be given an angle of launch theta.

My solution (as of now),

x = v cos θ .t -----------------------------------------------------------------Eqn 1
y = v sin θ + 0.5gt² (where g is -9.8, I will keep that in mind!)

We get the equation,

tan θ = (y-0.5gt²)/x

Solving for t,

t = √((x tanθ - y)/ (-0.5g)) (again g is -ve)

Now from eqn 1,

v = x / (cos θ.t)

Substituting for t,

v = (x) / [ cosθ . √((x tanθ - y)/(-0.5g)) ]

Now to find the stretch x,
k -> Spring constant, z-> stretch

Elastic Energy in spring = Kinetic energy

0.5kz² = 0.5mv²

z = √([mv² ] / k)

Substituting for v,

z = √[ (m/k)( (x²) / (cos²θ ((x tanθ - y)/(-0.5g)) ) ]

That was my grand equation. Let me know what you think.

 

[gneill]

Now, I'm confused. Let me explain the whole scenario.

I have to make a spring launcher that will fire a stretched spring. I will be given a target that is 'x' meters away and 'y' meters high w.r.t launch point. I will also be given an angle of launch theta.

My solution (as of now),

x = v cos θ .t -----------------------------------------------------------------Eqn 1
y = v sin θt + 0.5gt² (where g is -9.8, I will keep that in mind!)

We get the equation,

tan θ = (y-0.5gt²)/x

Solving for t,

t = √((x tanθ - y)/ (-0.5g)) (again g is -ve)

I think you'll find that:

t = \sqrt{\frac{2}{g}(y - x tan(\theta)}

Now from eqn 1,

v = x / (cos θ.t)

Substituting for t,

v = (x) / [ cosθ . √((x tanθ - y)/(-0.5g)) ]

Now to find the stretch x,
k -> Spring constant, z-> stretch

Elastic Energy in spring = Kinetic energy

0.5kz² = 0.5mv²

z = √([mv² ] / k)

Substituting for v,

z = √[ (m/k)( (x²) / (cos²θ ((x tanθ - y)/(-0.5g)) ) ]

That was my grand equation. Let me know what you think.

Is the spring compressed before or after the projectile is loaded onto it? The reason I ask is that there will be gravitational PE involvement in the compression/expansion of the spring.

 

[ChromoZoneX]

It is stretched before launch so that it can launch itself with the energy stored in it.

 

[ChromoZoneX]

Your expression for 't' is equivalent. I worked that out. :D

 

[gneill]

It is stretched before launch so that it can launch itself with the energy stored in it.

That is understood. But is the projectile loaded, the spring allowed to come to equilibrium, and THEN compressed by the required amount, or do you pre-compress the spring by the required amount and THEN place the projectile onto it?

 

[gneill]

Your expression for 't' is equivalent. I worked that out. :D

Suppose you try values: x = 30m ; y = 2m ; θ = 37° in each formula. Are the results the same?

 

[ChromoZoneX]

The spring is the projectile itself. It must be stretched at a given angle so that it can hit a target that is some horizontal distance away and some vertical distance high. I need to find how much it is stretched.

What is the solution you propose?

 

[gneill]

The spring is the projectile itself. It must be stretched at a given angle so that it can hit a target that is some horizontal distance away and some vertical distance high. I need to find how much it is stretched.

What is the solution you propose?

Hmm. I'm not sure what to suggest. The spring is going to have some mass (it is after all the mass of the projectile) and that mass is going to be distributed in some way along its length (maybe uniformly of you're lucky!). As it's stretched, the mass at different points on the spring will move different amounts in the gravitational field, making calculation of the gravitational PE a bit complicated.

If it's a very stiff spring the compression (or stretching) distance will be relatively small, so perhaps the gravitational PE will be negligible compared to the spring PE. That's something you can investigate.

Another question is how much of the spring's PE is going to end up as translational motion KE? The spring is essentially going to be performing an inelastic collision with itself, as the moving back end slams into the stationary front coil -- and then rebounds. I think that quite a bit of energy is going to end up making the spring continue to oscillate. The launch velocity may end up being just the velocity of the center of mass of the spring when it reaches its equilibrium length. I'm not sure about this though. A quick experiment with a spring might help.

 

[ChromoZoneX]

As far as I've been told, for our theoretical calculations, we're supposed to assume that all of the spring's elastic energy gets converted to kinetic energy. That is how I ended up with this,

mv^2 = kx^2

And finding 'x' as a result of all that other stuff in the thread.

 

[gneill]

As far as I've been told, for our theoretical calculations, we're supposed to assume that all of the spring's elastic energy gets converted to kinetic energy. That is how I ended up with this,

mv^2 = kx^2

And finding 'x' as a result of all that other stuff in the thread.

If that's what you're told to assume, then so be it. Proceed accordingly. You can find the required spring extension in the manner in which you've described