In 1780, in what is now referred to as \"Brady\'s Leap,\" Captain Sam Brady of t
ID: 1689879 • Letter: I
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, which is confined at that spot to a gorge. He landed safely on the far side of the river. It was reported that he leap 22 feet across while falling 20 feet.-Representing the horizontal distance jumped as L and the vertical drop as h , as shown in the figure, derive an expression for the minimum speed v he would need to make his leap if he ran straight off the cliff?
Explanation / Answer
No matter with what velocity he jumps off the leap, the vertical acceleration is always 9.8 m/s. Moreover his initial velocity while he is off the cliff is along the horizontal direction, say +ve x-axis. Hence he will take same time to fall 20 feet = 6.096 m given by s = ut + (1/2)*(9.8)*t^2 => 6.096 = 4.9 * t^2. => t = 1.115 sec. In this time he has to travel a horizontal distance of 22feet = 6.7056 m. Hence the velocity = 6.7056 / 1.115 = 6.0139 m/sec.