A person in an apartment building is 5.0 m
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?
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.