Rachel is on the roof of a building, \(h\) meters above ground. She throws a heavy ball into the air with a speed \(v,\) at an angle \(\theta\) with respect to the horizontal. Ignore air resistance. (a) Find the speed of the ball when it hits the ground in terms of \(h, v, \theta,\) and \(g .\) (b) For what value(s) of \(\theta\) is the speed of the ball greatest when it hits the ground?

First, we need to break down the initial velocity into horizontal and vertical components. To do this, we will use trigonometry: \(v_x = v \cos(\theta)\) (horizontal component) \(v_y = v \sin(\theta)\) (vertical component)

When the ball hits the ground, the vertical displacement is equal to the height of the building, \(h\). Using the initial vertical velocity \(v_y\), we will compute the time \(t\) it takes for the ball to hit the ground. Since the ball is moving upward first, \(h\) is positive: \(h = v_yt - \dfrac{1}{2}gt^2\) Solve for \(t\): \(t = \dfrac{v_y - \sqrt{v_y^2 - 2gh}}{g}\)

In order to find the speed of the ball when it hits the ground, we need to compute the horizontal displacement, \(x\), at this time. We can find the horizontal distance by multiplying the horizontal velocity by time: \(x = v_xt = (v\cos\theta)t\)

Now we need to find the vertical velocity before the ball hits the ground. We can find it using the equation: \(v_y^{\text{final}} = v_y - gt\)

Once we have the final vertical velocity, we can find the total velocity when the ball hits the ground. \(v^{\text{final}} = \sqrt{v_x^2 + (v_y^{\text{final}})^2}\) Substitute the values of \(v_x\) and \(v_y^{\text{final}}\) to find the total velocity in terms of the given parameters. \(v^{\text{final}} = \sqrt{(v\cos\theta)^2 + (v\sin\theta - gt)^2}\) Now we have the final velocity of the ball when it touches the ground.

To find the angle \(\theta\) for which the speed is greatest when the ball hits the ground, we need to maximize the expression of the final velocity. We can directly compute the derivative of the above expression given the associated complexity. Instead, we can use a simpler expression to find the angle maximizing the final velocity. It can be shown, using energy conservation, that the expression for the final speed is: \(v^{\text{final}} = \sqrt{v^2 + 2gh - 2vg\sin\theta}\) Now, we can differentiate this expression with respect to \(\theta\) and set the derivative equal to zero to find the angle maximizing the final speed: \(\dfrac{d(v^{\text{final}})}{d\theta} = \dfrac{dv\sin\theta}{d\theta}=0\) The above equation is satisfied when: \(\theta = 0 \text{ or }\sin\theta = 0\) Thus, to maximize the speed of the ball when it hits the ground, the angle \(\theta\) should be \(0^{\circ}\). This means that when the ball is thrown horizontally, its speed will be maximum when it reaches the ground.