# Finding the initial velocity and time period of a projectile

1. Feb 25, 2015

### Wisal

1. The problem statement, all variables and given/known data

In the Fig.
What should be the initial velocity (magnitude and direction) of the arrow? Also calculate the time of flight.

2. Relevant equations

3. The attempt at a solution
I have formulated 2 equations:
v*cos(theta) * t = 20
and 3 = (v * Sin(theta) * t) + (9.81 * t^2)/2
when t = time of flight
and v = initial velocity

I have further gone to a final equation
v = (2000+2000(tan(theta))^2)/ (3 + 20 Tan(theta))

2. Feb 25, 2015

### PeroK

Do you think there is only one way to hit the apple?

3. Feb 25, 2015

### Wisal

Well, intuitively speaking, I think that this problem will have a range of values as we can hit the apple along a variety of trajectories.

4. Feb 25, 2015

### PeroK

That's correct. So, could you clarify what you are trying to calculate? For example, there will be a minimum value of v for which you can hit the apple.

5. Feb 25, 2015

### Wisal

But won't there be an optimal solution to this, the solution that takes the least time? I mean the value of v that takes the least time?

6. Feb 25, 2015

### PeroK

There's not really going to be a least time, as the least time will be for the fastest v.

I think there would be a longest time, associated with the lowest possible velocity.

7. Feb 25, 2015

### Wisal

Oh yes! How stupid of me to ask :P But how will we find the lowest value of v?

8. Feb 25, 2015

### PeroK

Are you sure you are not missing a constraint from the problem? It's quite tricky to calculate the lowest v to hit a point above the ground.

9. Feb 25, 2015

### Wisal

well the problem, unfortunately, in front of me only gives this information. I think that there will be different minimum velocity for every possible angle. Am I right? and how do we find the range of angles possible?

10. Feb 25, 2015

### PeroK

I'm pretty sure it's not the problem you're supposed to be doing, but if you want a challenge.

First step is to express the trajectory as a (parabolic) function. Here's the formula. Try to derive this:

$y = xtan\theta - \frac{gx^2}{2v^2}sec^2\theta$

11. Feb 25, 2015

### Wisal

I have already done that:
I formulated two equations:
1. vCosθ * t = 20
as x =20
2. 3 = vSinθ(t) + g(t^2)/2
as y =3 and then substituting the value of 't' i.e. 20/vcosθ in the second equation

12. Feb 25, 2015

### PeroK

Okay, but remember g = -9.81 in your equations.

If you're trying to minimise v, then you need to maximise $\theta$ which is the same as maximising $tan\theta$. Note that $sec^2\theta = tan^2\theta + 1$

So, aim for a quadratic equation in $tan\theta$.

13. Feb 25, 2015

### Wisal

Okay, that's what I have reached at:

is the answer 89.7 degrees?

14. Feb 25, 2015

### Wisal

And also could you tell me the intuitive reason why we took g as -9.81( I am sometimes confused by this). I do get why we took this value in this particular problem as we had to get a maximum value of tanθ hence we get a parabola with a maximum point.

And, am I right in thinking that there will be a different minimum velocity for every possible angle?

15. Feb 25, 2015

### PeroK

That can't be correct.

16. Feb 25, 2015

### Wisal

y = xtanθ - (gx^2 secθ^2)/2 Comment: as we take g =- 9.81 so negative sign)
2y = 2xtanθ-gx^2secθ^2
2y = 2xtanθ-gx^2-gx^2(tanθ)^2
2y+gx^2=2xtanθ-gx^2(tanθ)^2
now plugging in the values: x = 20; y =3
2(3) +(9.81*400)=2(20)tanθ-(9.81*400)(tanθ)^2
now by substituting any variable say 'x' as tanθ
-3924x^2 + 40 x - 3930 = 0
Then differentiating it and putting that to zero and getting the value of x to be 196.2 and then using tanθ = 196.2 and then finding out the value of θ.

17. Feb 25, 2015

### PeroK

I might have led you astray! We need to maximise the quadratic function for a given v. And, if the maximum value is y, we have the minimum value for v. It's not maximising theta. We need to maximise:

$f(\theta) = xtan\theta - \frac{gx^2}{2v^2}(1 + tan^2\theta)$

I told you it was tricky!

18. Feb 25, 2015

### Wisal

But is not the 'y'(vertical displacement) always constant as the displacement will always be 3 no matter what? I mean the final displacement will be 3. Thanks for the help :D

19. Feb 25, 2015

### PeroK

That's not the point. Suppose we fire the arrow at a velocity v. That function gives the height at x for a given $\theta$. So, first we calculate the maximum height at x for a given v (for any $\theta$).

If that maximum height happens to be y (=3 in this case), then we have found the minumum v. Think about it. If we could hit (x, y) with a lower velocity than v, then we could hit a higher point (x, something higher than y) at velocity v.

In other words, at the minimum v, we can only just hit the target. And no higher.

Last edited: Feb 25, 2015