Question

In 1780, in what is now referred to as "Brady's Leap," Captain Sam Brady of the U.S. Continental Army escaped certain death from his enemies by running over the edge of the cliff above Ohio's Cuyahoga River in (Figure 1) , which is confined at that spot to a gorge. He landed safely on the far side of the river. It was reported that he leapt 22 ft (≈ 6.7 m) across while falling 20 ft (≈ 6.1 m).
What is the minimum speed with which he’d need to run off the edge of the cliff to make it safely to the far side of the river?

Express your answer to two significant figures and include the appropriate units.

Answer 1

The minimum speed with which Captain Brady had to run off the edge of the cliff to make it safely to the far side of the river is around 6 meters per second.

Further explanation

This is a free fall 2-dimensional type of problem, therefor we can write equations for both dimensions which model the fall of captain Brady. Let's call x the distance travelled by the captain on the horizontal direction and y the distance travelled on the vertical direction.

Lets suppose that Brady jumped with a complete horizontal velocity from a point which we will call the origin (meaning zero horizontal and vertical displacement), and let's call ta the time it took for captain Brady to reach the river (meaning the time he spent on the air). The equations of motion for the captain will be:

$x= V \cdot t$

$y= - \frac{g \cdot t^2}{2}$

We know that at time ta the captain would have traveled 6.7 m on the horizontal direction, and 6.1 m in the vertical direction. Therefor we can write that:

$6.7= V \cdot ta$

$-6.1= - \frac{g \cdot {ta}^2}{2}$

Which gives us a system of 2 equations and 2 unknowns (V and ta). From the second equation we can solve for ta as:

$ta = \sqrt{\frac{2 \cdot 6.1}{g}} =1.12 s$

And solving for V on the first equation, we find that:

$V= \frac{6.7}{1.12} = 5.98 \frac{m}{s}$

Which is almost 6 meters per second.

Learn more

• Free fall of an arrow: brainly.com/question/1597396
• Concept of free fall: brainly.com/question/1708231

Keywords

Free fall, projectile, gravity

Answer 2

The minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$.

Further explanation:

As Captain Sam Brady jumps from the cliff, he moves in two dimension under the action of gravity.

Given:

The height of free fall of the captain Brady is $20\text{ ft}$ or $6.1\text{ m}$.

The horizontal distance moved by the captain Brady is $22\text{ ft}$ or $6.7\text{ m}$.

Concept:

The time required to free fall of a body can be calculated by using the expression given below.

$\left( { - s}\right)=ut-\frac{1}{2}g{t^2}$                                 ……. (1)

The displacement is considered negative because the captain is moving in vertically downward direction.

Here, $s$ is the distance covered by the body in free fall, $u$ is the initial velocity of the object, $g$ is the acceleration due to gravity and $t$ is the time taken in free fall of a body.  

As the Caption jumps off the cliff, he has his velocity in the horizontal direction. The velocity of the captain in vertical direction is zero.

Substitute $0$ for $u$ in the equation (1) .

$s=\frac{1}{2}g{t^2}$  

Rearrange the above expression for $t$.

$\boxed{t=\sqrt {\frac{{2s}}{g}}}$                                                              …… (2)  

Converting acceleration due to gravity in $\text{ft}/\text{s}^2$ .

$\begin{aligned}g&=\left( {9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\left( {\frac{{1.0\,{\text{ft/}}{{\text{s}}^{\text{2}}}}}{{0.305\,{\text{m/}}{{\text{s}}^{\text{2}}}}}} \right) \\&=32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}} \\ \end{aligned}$

Substitute $20\text{ ft}$ for $s$ and $32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}$ for $g$ in equation (2) .

$\begin{aligned}t&=\sqrt {\frac{{2\left( {20\,{\text{ft}}} \right)}}{{\left( {32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}} \right)}}} \\&=1.116\,{\text{s}} \\ \end{aligned}$

Therefore, the time taken by captain to free fall a height $20\text{ ft}$ is $1.116\text{ s}$.  

In the same time interval captain has to move $22\text{ ft}$ in horizontal direction. The acceleration is zero in horizontal direction. So, the velocity will be constant throughout the motion in the horizontal direction.

The distance travelled by captain in the horizontal direction is given by,

$x=v\cdot t$

Rearrange the above expression for $v$.  

$\boxed{v=\dfrac{x}{t}}$                                                      …… (3)  

Here, $x$ is the distance travelled in horizontal direction, $v$ is the velocity of the captain and $t$ is the time.  

Substitute $22\text{ ft}$ for $x$ and $1.116\text{ s}$ for $t$ in equation (3) .

$\begin{aligned}v&=\frac{{22\,{\text{ft}}}}{{1.116\,{\text{s}}}} \\&=19.71\,{\text{ft/s}} \\ \end{aligned}$

Thus, the minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$.

Learn more:

1. Energy density stored in capacitor brainly.com/question/9617400

2. Kinetic energy of the electrons brainly.com/question/9059731

3. Force applied by the car on truck brainly.com/question/2235246

Keywords:

Free fall, projectile, gravity, 1780, Brady’s, leap, Captain, Sam Brady, US, continental army, enemies, Ohio’s, Cuyahoga river, 22 ft, 6.7 m, 20 ft, 6.1 m, minimum speed, run off, edge, cliff, safely, far side, river, 19.71 ft/s, 6 m/s, 6 meter/s, 5.99 m/s, 599.8 cm/s.