A person in an apartment building is 5.0 m

Question

above a person walking below. She plans
to drop some keys to him. He is currently
walking directly toward a point below her
at 2.75 m/s. How far away is he if he
catches the keys 1.25 m above the ground?

Answer 1

To solve this problem, we can use the physics equation for the motion of an object in free fall:

h = (1/2)gt^2

where h is the vertical distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

First, let's find the time it takes for the keys to fall 5.0 m. Using the equation above, we have:

5.0 = (1/2)(9.8)t^2

Rearranging the equation, we get:

t^2 = (2)(5.0) / 9.8

t^2 = 1.02

Taking the square root of both sides, we find:

t = √(1.02) = 1.01 seconds

Now, let's find how far the person walking below travels in 1.01 seconds. The distance traveled is given by:

d = vt

where d is the distance, v is the velocity, and t is the time.

In this case, the velocity is 2.75 m/s and the time is 1.01 seconds:

d = (2.75)(1.01) = 2.7775 meters

Therefore, the person is approximately 2.7775 meters away from the point directly below the person in the apartment building when he catches the keys 1.25 m above the ground.