Question

You are standing at the edge of a cliff which is 40 m tall. You throw an apple up with a speed of 15 m/s, and it accelerates downward at 9.81 m/s^2. How long does it take to fall to the bottom of the cliff

Answer 1

Answer:

Approximately $4.77\; {\rm s}$.

Explanation:

Let $u$ denote the initial velocity of this apple. Let $v$ denote the velocity of the apple right before landing (final velocity.) Let $x$ denote the displacement of this apple (from the edge to the bottom.) Let $a$ denote the acceleration of this apple. Let $t$ denote the time it takes for the apple to land.

The acceleration of this apple is $a = (-9.81)\; {\rm m\cdot s^{-2}}$. This value is negative since the apple is accelerating downwards.

It is given that the initial velocity of the apple was $u = 15\; {\rm m\cdot s^{-1}}$. Note that unlike $a$, the value of $u$ is positive since the apple was initially travelling upwards.

The displacement of the apple would be $x = (-40)\; {\rm m}$- equal to the height of the cliff in magnitude, but negative since the apple would land at a location below the edge.

Since the acceleration of this apple is a constant value, the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ will apply.

Rearrange this equation and solve for $v$ (velocity of apple right before landing):

$\begin{aligned}v^{2} &= 2\, a\, x + u^{2}\end{aligned}$.

Note that the apple will be travelling downward right before it lands. Therefore, the value of $v$ (velocity right before the apple lands) will be negative:

$\begin{aligned}v &= -\sqrt{2\, a\, x + u^{2}}\end{aligned}$.

Substitute in $a = (-9.81)\; {\rm m\cdot s^{-2}}$, $x = (-40)\; {\rm m}$, and $u = 15\; {\rm m\cdot s^{-1}}$:

$\begin{aligned}v &= -\sqrt{2\, a\, x + u^{2}} \\ &= -\sqrt{2\times (-9.81)\; {\rm m\cdot s^{-2}}\times (-40)\; {\rm m} + (15\; {\rm m\cdot s^{-1}})^{2}} \\ &\approx -31.78\; {\rm m\cdot s^{-1}}\end{aligned}$.

In other words, the velocity of the apple would have changed from $u = 15\; {\rm m\cdot s^{-1}}$ to $(-31.78)\; {\rm m\cdot s^{-1}}$ during the flight. The velocity change would be:

$\begin{aligned} \Delta v &= v - u \\ &\approx (-31.78)\; {\rm m\cdot s^{-1}} - 15\; {\rm m\cdot s^{-1}} \\ &= -46.78\; {\rm m\cdot s^{-1}}\end{aligned}$.

At a rate of $a = (-9.81)\; {\rm m\cdot s^{-2}}$, the time it takes to achieve such velocity change would be:

$\begin{aligned} t &= \frac{\Delta v}{a} \\ &\approx \frac{(-46.78)\; {\rm m\cdot s^{-1}}}{(-9.81)\; {\rm m\cdot s^{-2}}} \\ &\approx 4.77\; {\rm s} \end{aligned}$.