Physics, 19.08.2019 14:30 khuak877

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.

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Answer from: james22000

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, is the distance covered by the body in free fall, is the initial velocity of the object, is the acceleration due to gravity and 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 for in the equation (1) .

$s=\frac{1}{2}g{t^2}$

Rearrange the above expression for .

$\boxed{t=\sqrt {\frac{{2s}}{g}}}$ …… (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 and $32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}$ for 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 .

$\boxed{v=\dfrac{x}{t}}$ …… (3)

Here, is the distance travelled in horizontal direction, is the velocity of the captain and is the time.

Substitute $22\text{ ft}$ for and $1.116\text{ s}$ for 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

2. Kinetic energy of the electrons

3. Force applied by the car on truck

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.

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