The determined Wile E. Coyote is out once

(a) Derive the expression for the minimum time required for the coyote to travel a distance of $\mathrm{\Delta }x$ from the cliff.

The distance travelled by the coyote is,

$\mathrm{\Delta }x=v_{0x}t+\frac{1}{2}{a}_x{t}^2$

Here $v_{0x}$ is the initial horizontal velocity and $a_x$ is the horizontal acceleration of the coyote, and $t$ is the time to cover the distance $\mathrm{\Delta }x$.

The initial velocity og the coyote is zero.

$v_{0x}=0m/s$

Substitute $0m/s$ for $v_{0x}$ in the equation $\mathrm{\Delta }x=v_{0x}t+\frac{1}{2}{a}_x{t}^2$ and solve for $t$.

$\mathrm{\Delta }x=(0m/s)t+\frac{1}{2}{a}_x{t}^2$

$t=\sqrt{\frac{2\mathrm{\Delta }x}{a_x}}$

The roadrunner is moving with a constant speed. Hence, the minimum speed to reach the cliff is as follows.

$v_{min}=\frac{\mathrm{\Delta }x}{t}$

Substitute $\sqrt{\frac{2\mathrm{\Delta }x}{a_x}}$ for $t$.

$v_{min}=\frac{\mathrm{\Delta }x}{\sqrt{\frac{2\mathrm{\Delta }x}{a_x}}}$

$=\sqrt{\frac{a_x\mathrm{\Delta }x}{2}}$

Substitute $15m/s^2$ for $a_x$ and $70m$ for $\mathrm{\Delta }x$

$v_{min}=\sqrt{\frac{(15m/s^2)(70m)}{2}}$

$=23m/s$

Hence, the required minimum speed of the roadrunner is $23m/s$

b) The initial velocity of the coyote when it goes over the cliff is equal to the horizontal velocity.

$v_{0x}=a_{x}t$

Substitute $\sqrt{\frac{2\mathrm{\Delta }x}{a_x}}$ for $t$.

$v_{0x}=a_x\sqrt{\frac{2\mathrm{\Delta }x}{a_x}}$

$=\sqrt{2a_x\mathrm{\Delta }x}$

The time taken to cover the vertical height of the cliff is,

$t^{\prime }=\sqrt{\frac{2\mathrm{\Delta }y}{a_y}}$

Here $\mathrm{\Delta }y$ is vertical heisght of the cliff.

Therefore, the horizontal displacement of the fall of the coyote is,

$\mathrm{\Delta }x=v_{0x}t+\frac{1}{2}{a}_x{t}^2$

Substitute  $=\sqrt{2a_x\mathrm{\Delta }x}$ for v_{0x} and $\sqrt{\frac{2\mathrm{\Delta }y}{a_y}}$ for $t^{\prime }$.

${\displaystyle \mathrm{\Delta }x=\sqrt{2a_x\mathrm{\Delta }x}(\sqrt{\frac{2\mathrm{\Delta }y}{a_y}})+\frac{1}{2}{a}_x(\sqrt{\frac{2\mathrm{\Delta }y}{a_y}}{)}^2}$

 Substitute $15m/s^2$ for $a_x$ and $70m$ for $\mathrm{\Delta }x$, $-100m$ for $\mathrm{\Delta }y$, and $-9.80m/s^2$ for $a_y$

${\displaystyle \mathrm{\Delta }x=\sqrt{2(15m/s)(70m)}(\sqrt{\frac{2(-100m)}{-9.80m/s^2}})+\frac{1}{2}(15m/s)(\sqrt{\frac{2(-100m)}{-9.80m/s^2}}{)}^2}$

$=3.6\times {10}^{2}m$

Hence, the reauired dispplacement is $3.6\times {10}^{2}m$