Question

A person standing at the top of a hemispherical rock of radius $\mathit{R}$kicks a ball (initially at rest on the top of the rock) to give it horizontal velocity ${\overrightarrow{\mathbf{v}}}_{\mathbf{i}}$ as shown in Figure P.484. (a) What must be its minimum initial speed if the ball is never to hit the rock after it is kicked? (b) With this initial speed, how far from the base of the rock does the ball hit the ground?

Short answer

(a) The minimum speed must be $\sqrt{Rg}$.

(b) The distance from the base of the stone is $.414R$.

Step-by-step solution

Definition of downstream.

Downstream refers to the direction of or proximity to a stream's mouth.

Find the minimum initial speed if the ball is never to hit the rock after it is kicked.

(a)

Let’s consider the minimum initial speed must be $v$.

$\begin{array}{l}x=vt\\ y=R-\frac{1}{2}g{t}^2\end{array}$

Therefore,

$t=\frac{x}{v}$

Substitute the value of $t$,

$y=R-\frac{1}{2}g{\left(\frac{x}{v}\right)}^2$

Now, the equation of circle is given by

$\begin{array}{c}{x}^2+y^2=R^2\\ y^2=R^2-x^2\end{array}$

The $y$from the projectile is always bigger than the $y$ from the circle since the ball never strikes the stone.

Hence,

$\begin{array}{c}{\left[R-\frac{1}{2}g{\left(\frac{x}{v}\right)}^2\right]}^2>R^2-x^2\\ R^2+\frac{g^2{x}^4}{4v^2}-\frac{Rg{x}^2}{v^2}>R^2-x^2\\ \frac{g^2{x}^4}{4v^2}-\frac{Rg{x}^2}{v^2}>-x^2\\ \frac{g^2{x}^2}{4v^2}-\frac{Rg}{v^2}>-1\end{array}$

Now, even if $x=0$, the following equation holds true for any arbitrary value of $x$. Hence,

$\begin{array}{c}-\frac{Rg}{v^2}>-1\\ \frac{Rg}{v^2}<1\\ v^2>Rg\\ v>\sqrt{Rg}\end{array}$

Hence, the minimum speed must be $\sqrt{Rg}$, if the ball never touches the stone.

Find the distance from the base of the rock does the ball hit the ground.

(b)

The time taken by ball to reach ground is

$\begin{array}{c}t=\sqrt{\frac{2h}{g}}\\ =\sqrt{\frac{2R}{g}}\end{array}$

Hence, the horizontal distance travelled by the ball is

$\begin{array}{c}d=vt\\ =\sqrt{Rg}\sqrt{\frac{2R}{g}}\\ =\sqrt{2}R\end{array}$

Hence, the distance from the base of the stone is

$\begin{array}{c}d\text{'}=d-R\\ =\sqrt{2}R-R\\ =0.414R\end{array}$