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? the answer was 6 m/s the world-record time for the 100 m dash is approximately 10 s. given this, is it reasonable to expect brady to be able to run fast enough to achieve brady's leap?

a.yes, the obtained speed is less than the world-record.

b.yes, the obtained speed is almost equal to the world-record.

c.no, the obtained speed is greater than the world-record.

d.no, the obtained speed is almost equal to the world-record.

-Samuel Brady gained his lasting notoriety for his leap over the Cuyahoga River around 1780 in what is now Kent, Ohio. After following a band of Indians into the Ohio country, a failed ambush attempt resulted in the band chasing Brady near the Cuyahoga River. To avoid capture, Brady leaped across a 22-foot (6.7 m) wide gorge of the river (which was widened considerably in the 1830s for construction of the Pennsylvania and Ohio Canal) and fled to a nearby lake where he hid in the water under a fallen tree using a reed for air.

adityaso adityaso · Answer 2 · 2019-08-28T05:11:55-04:00

Part (a): The minimum speed of Brady should be \(\boxed{6\text{ m/s}}\) to cross the cliff.

Part (b): The speed of Brady’s leap is possible to achieve, as it is less than the speed of world record.

Further Explanation:

(a)

Brady jumps from a cliff to go on other side. He crosses \(22\text{ ft}\) horizontal distance from a height of \(20\text{ft}\). To jump from cliff, he follows the Newton’s law of motion.

Given:

The horizontal distance is \(22\text{ ft}\).

The vertical distance is \(20\text{ft}\).

Concept:

The horizontal distance in meter is \(6.71\text{ m}\).

The vertical distance in meter is \(6.10\text{ m}\).

To obtain the time of flight, applying one of the equation of motion given as:

\(s=ut+\dfrac{1}{2}at^2\)

Substitute \(0\text {m/s}\) for \(u\) and rearrange the above equation for \(t\) :

\(t=\sqrt{\dfrac{2s}{a} }\)

Substitute \(6.10\text{ m}\) for \(s\) and \(9.81\text{ m}/\text{s}^2\) for \(a\) in above equation.

\(\begin{aligned}t&=\sqrt{\dfrac{2\times6.10}{9.81}}\text{ s}\\&=1.12\text{ s}\end{aligned}\)

The time of flight and time taken to cover horizontal distance are equal.

Horizontal velocity of Brady given as:

\(\begin{aligned}v&=\dfrac{6.71}{1.12}\text{ m/s}\\&\approx6\text{ m/s}\end{aligned}\)

Thus, the minimum speed of Brady should be \(\boxed{6\text{ m/s}}\) to cross the cliff.

(b)

The minimum speed with which Brady is running is \(6\text{ m/s}\).

The speed of world record given as:

\(\begin{aligned}V&=\frac{100}{10}\text{ m/s}\\&=10\text{ m/s}\end{aligned}\)

Thus, the speed of Brady’s leap is possible to achieve, as it is less than the speed of world record.

Learn more:

1. Projectile motion of a body: https://brainly.com/question/11023695

2. Body in pure rolling motion: https://brainly.com/question/9575487

3. Newton’s law of motion: https://brainly.com/question/6125929

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

1780, Brady's Leap, Captain, Sam, U.S. Continental Army, horizontally, cliff, Ohio's Cuyahoga, gorge, leap, 22 ft, 20 ft, minimum, speed, river, 100 m, dash, 10 s, jump, time of flight, vertical and distance.