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?  

Answer 1

Answer 2

Calculation Steps:

1. Understand the Problem:
   Brady leaps horizontally 6.7 m (22 ft) while falling vertically 6.1 m (20 ft). We

2. need to find his initial horizontal speed.

3. 
   

4. Key Physics Concepts:

• Horizontal motion: Constant speed (no acceleration).

• Vertical motion: Free-fall under gravity ($a=g=9.8{\text{m/s}}^2$).

• 
  

5. Time in the Air:
   First, calculate how long it takes Brady to fall 6.1 m vertically using the free-fall equation:

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

   Solve for time ($t$):

   $$6.1=\frac{1}{2}\times 9.8\times t^2⟹t^2=\frac{6.1\times 2}{9.8}\approx 1.24⟹t\approx \sqrt{1.24}\approx 1.11\text{s}$$

7. Horizontal Speed:
   To cover 6.7 m horizontally in 1.11 s, his initial speed ($v_x$) must be:

   $$v_x=\frac{\text{Horizontal distance}}{\text{Time}}=\frac{6.7}{1.11}\approx 6.4\text{m/s}$$

9. Answer:
   The minimum speed Captain Sam Brady needed was approximately 6.4 m/s.