Question

A boy throws a stone horizontally from the top of a cliff of height $\mathit{h}$toward the ocean below. The stone strikes the ocean at distance$\mathit{d}$from the base of the cliff. In terms of$\mathit{h}\mathbf{,}\mathit{d}$, and$\mathit{g}$, find expressions for

(a) the time$\mathit{t}$at which the stone lands in the ocean,

(b) the initial speed of the stone,

(c) the speed of the stone immediately before it reaches the ocean, and

(d) the direction of the stone's velocity immediately before it reaches the ocean.

Short answer

(a) The landing time of the stone in the ocean is$\sqrt{\frac{2h}{g}}$.

(b) The horizontal speed of the stone is$d\sqrt{\frac{g}{2h}}$.

(c) The speed of stone just before reaching the ocean is$\sqrt{\frac{d^{2}g}{2h}+2gh}$.

(d) The direction of the stone's velocity before reaching the ocean is ${\tan }^{-1}\left(\frac{2h}{d}\right)$.

Step-by-step solution

 Step 1: The vertical displacement of the stone.

The vertical displacement of the stone is given by

$\mathbf{-}\mathit{h}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{g}{\mathit{t}}^{\mathbf{2}}$

Here,$\mathit{h}$is the vertical displacement,$\mathit{g}$is the acceleration due to gravity and$\mathit{t}$is the time required for the stone to land in the ocean.

 Step 2: The landing time of the stone in the ocean.

(a)

The height of the cliff is$h$. The distance between the base of the cliff and the position where the stone strikes the ocean is$d$.

The initial vertical velocity is zero.

Write the formula to calculate vertical displacement of the stone

$-h=-\frac{1}{2}g{t}^2$

Here, $h$is the vertical displacement, $g$is the acceleration due to gravity and $t$is the time required for the stone to land in the ocean.

Rearrange to the above equation to find$t$.

$t=\sqrt{\frac{2h}{g}}$

Therefore, the landing time of the stone in the ocean is $\sqrt{\frac{2h}{g}}$.

The initial speed of the stone.

(b)

In the horizontal direction there is no acceleration.

The formula to calculate initial speed of the stone

$v_{\mathrm{xi}}=\frac{d}{t}$

Here,$v_{\mathrm{xi}}$ is the initial speed of the stone in $x$direction and $d$is the distance covered by stone.

Substitute $\sqrt{\frac{2h}{g}}$for $t$to find .

$\begin{array}{c}{v}_{\mathrm{xi}}=\frac{d}{\sqrt{\frac{2h}{g}}}\\ =d\sqrt{\frac{g}{2h}}\end{array}$

Therefore, the initial speed of the stone is $d\sqrt{\frac{g}{2h}}$.

The speed of stone before it reaches the ocean.

(c)

Write the formula to calculate final speed of the stone

$v_{\mathrm{y}}^2=u_{\mathrm{y}}^2+2gh$

Here,$v_y$is the final speed of the stone in$y$direction and$u_y$is the initial speed of the stone in$y$direction.

Substitute 0 for$u_y$to find$v_y$.

$\begin{array}{l}{v}_{\mathrm{y}}^2=0^2+2gh\\ v_{\mathrm{y}}=\sqrt{2gh}\end{array}$

The formula to calculate speed of the stone before reaching the ocean

$v=\sqrt{v_{\mathrm{xi}}^2+{v_{\mathrm{y}}}^2}$

Here,$v$is the final speed of the stone.

Substitute$\sqrt{2gh}$for$v_y$and$d\sqrt{\frac{g}{2h}}$for$v_{xi}$to find$v$.

$\begin{array}{c}v=\sqrt{{\left(d\sqrt{\frac{g}{2h}}\right)}^2+{\left(\sqrt{2gh}\right)}^2}\\ =\sqrt{\frac{d^{2}g}{2h}+2gh}\end{array}$

Therefore, the speed of stone before it reaches the ocean is $\sqrt{\frac{d^{2}g}{2h}+2gh}$.

The direction of the stone's velocity before reaches the ocean.

Write the formula to calculate direction of the stone's velocity$\tan \theta =\frac{v_y}{v_{xi}}$

Here, $\theta$is the direction of stone's velocity.

Substitute$\sqrt{2gh}$for$v_y$and$d\sqrt{\frac{g}{2h}}$for$v_{xi}$to find$\theta$.

$\begin{array}{l}\tan \theta =\frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}\\ \theta ={\tan }^{-1}\left(\frac{2h}{d}\right)\end{array}$

Therefore, the direction of the stone's velocity before reaches the ocean is${\tan }^{-1}\left(\frac{2h}{d}\right)$.