A man stands on the edge of a cliff with hands outstretched horizontally
carrying identical pebbles, one in each hand. At time t= 0, he drops the first pebble from rest.When this pebble has fallen by 2.0 m, he releases the second pebble. What is the vertical distance between the pebbles at t= 5s?
(You may neglect air resistance and assume that neither pebble
has hit the ground at 5.0 s.)
Particle 1
v = g t
d = (1/2) g t^2
how long to fall 2 meters?
4 = 9.81 t^2
t = .408 seconds
Particle 2
v' = g (t-.408)
d' = (1/2) g t(t -.408)^2
at t = 5
d = (1/2)(9.81)(25)
d' = (1/2)(9.81)(4.59)^2
so calculate d'-d
t is not 0.408. t^2 is 0.408. and t=0.63 s for 2nd h=5s-0.63s=4.37s
h=1/2gt^2
h1=1/2x9.81x(5)^2=123 m
h2=1/2x9.81x(4.37)^2=95.48 m
h1-h2=27.145 m
To find the vertical distance between the pebbles at t=5s, we need to determine the height of each pebble at that time. We can use the equation of motion for freely falling objects to solve this problem.
The equation we can use is:
h = (1/2) * g * t^2
Where:
h is the height
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time
Let's calculate the height of each pebble at t=5s:
For the first pebble:
h1 = (1/2) * g * t^2
= (1/2) * 9.8 * (5^2)
= (1/2) * 9.8 * 25
= 122.5 meters
For the second pebble, it is dropped after the first pebble has fallen by 2.0 m. Therefore, the height of the second pebble at t=5s will be:
h2 = h1 - 2.0
= 122.5 - 2.0
= 120.5 meters
So, the vertical distance between the pebbles at t=5s is 120.5 meters.