51.3k views
1 vote
A stunt person jumped from the roof of one building to the balcony of another building 5 meters below. The distance between the 2 buildings is 3 meters. What is the minimum horizontal velocity the stunt person must have to accomplish this jump (ignore air resistance)?

a) 10 m/s
b) 15 m/s
c) 20 m/s
d) 25 m/s

1 Answer

1 vote

Final answer:

To accomplish the jump between the two buildings, the stunt person must have a minimum horizontal velocity of approximately 2.94 m/s.

Step-by-step explanation:

To calculate the minimum horizontal velocity the stunt person must have to accomplish the jump, we can use the concept of projectile motion. The horizontal distance between the two buildings is 3 meters, and the vertical distance is 5 meters. We can use the formula v = d / t where v is the horizontal velocity, d is the horizontal distance, and t is the time taken.

First, we need to find the time taken for the object to fall 5 meters vertically. We can use the formula h = (1/2)gt² where h is the vertical distance, g is the acceleration due to gravity (9.8 m/s²), and t is the time taken. Plugging in the values, we get 5 = (1/2)(9.8)t². Solving for t, we find t ≈ 1.02 s.

Next, we can substitute the value of t into the formula v = d / t. Plugging in the values, we get v = 3 / 1.02 ≈ 2.94 m/s. Therefore, the minimum horizontal velocity the stunt person must have to accomplish the jump is approximately 2.94 m/s.

User Kaptein Babbalas
by
8.8k points