# Finding the Minimum θ for Rock Thrown at Speed v_0

• darksyesider
In summary, the minimum value of θ, in terms of h and v_0, can be found by differentiating the tangent of the equation v_x = v_0cosα and setting it equal to zero. This will eliminate α and result in the equation v_0^2sin^2α = 2gh.

## Homework Statement

A boy throws a rock at speed v_0 at angle α from a balcony of height h. When the ball hits the ground, its velocity makes an angle θ with the ground. What is the minimum value of θ in terms of h and v_0

## Homework Equations

basic kinematics equations.

## The Attempt at a Solution

I started out with the following:

$$v_x = v_0 \cos \alpha , v_y = \sqrt{v_0^2\sin^2 \alpha+2gh}$$

Then:

$$f( \theta ) = \tan^{-1} (\frac{\sqrt{v_0^2 \sin^2 \alpha+2gh}}{v_0\cos\alpha}$$

To find the minimum, you have to differentiate, although whenever I do this i get a HUGE expression which doesn't make sense. Any help?Thanks :)

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When ## \theta ## is max, ## \tan \theta ## is max as well.

But we need to minimize theta?
Can you please give another hint?

Ah, sorry. I really meant to say min, not max :)

Gotcha.

So am I correct in saying that after taking the tangent of both sides, the RHS equals zero? in which case :

$$v_0^2 \sin^2 \alpha = 2gh$$ ??

I attempted it again, however i cannot get rid of alpha here.

You need ## (\tan \theta)' = 0 ##.

I still don't get it. Why is that so? Thanks.

Let ## f(x) ## be a monotonic function. Let ## g(x) ## be any function. Consider ## F(x) = f(g(x)) ##. ## F'(x) = (f(g(x)))' = f'(g(x)) g'(x) = 0 ##. Because ## f(x) ## is monotonic, ## f'(x) \ne 0 ##, thus the previous equation implies ## g'(x) = 0 ##, which means that if ## F(x) ## has a stationary point, ## g(x) ## has a stationary point there as well.

## \tan ## is monotonic from zero to ## \pi/2 ##.

So does this mean I have to differentiate $$\frac{\sqrt{v_0^2\sin^2\alpha + 2gh}}{v_0\cos\alpha}$$?
Also, what do i differentiate with respect to? And how do i get rid of alpha? sorry I'm very stuck on this

Since you are required to find that in terms of ##h## and ##v_0##, ##\alpha## has to disappear, which apparently means the minimum must be with respect to it.

## 1. What is the formula for finding the minimum θ for a rock thrown at speed v0?

The formula for finding the minimum θ is θ = arctan(v02/gR), where v0 is the initial speed of the rock, g is the acceleration due to gravity, and R is the radius of the circular path the rock must follow.

## 2. How does the initial speed of the rock affect the minimum θ?

The initial speed of the rock, v0, directly affects the minimum θ. As v0 increases, the minimum θ decreases. This means that the rock must be thrown at a higher initial speed in order to achieve a smaller minimum angle.

## 3. What is the significance of the minimum θ in this scenario?

The minimum θ represents the angle at which the rock must be thrown in order to achieve the farthest distance. Any angle above the minimum θ will result in a shorter distance, while any angle below the minimum θ will result in the rock falling short of the desired distance.

## 4. Can the minimum θ be negative or greater than 90 degrees?

No, the minimum θ cannot be negative or greater than 90 degrees. This is because the minimum θ is the angle at which the rock must be launched in order to achieve the farthest distance, and launching the rock at a negative angle or an angle greater than 90 degrees would not result in a valid trajectory.

## 5. Does the minimum θ change if the radius of the circular path is altered?

Yes, the minimum θ will change if the radius of the circular path is altered. As the radius increases, the minimum θ will also increase. This is because a larger radius requires a greater angle in order for the rock to travel the same distance.