# Throwing a ball

## 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 :)

#### Attachments

• Screen Shot 2013-09-14 at 11.18.41 AM.png
6.5 KB · Views: 350
Last edited:

## Answers and Replies

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.