an object released from rest will fall 16t^2 feet, t seconds after it is dropped. 2 pebbles are dropped, one after the other, from a cliff that is 64 feet high. the 2nd pebble is released the instant the first pebble has fallen exactly one foot. how far above the ground is the second pebble when the first pebble strikes the ground?
1st pebble falls 1 foot in 1/4 sec
1st pebble takes 2 seconds to hit
during 7/4 seconds, second pebble falls 49 feet
so it is still 11 feet up when the first pebble hits
is there an equation for this?
I don,t know that what I am trying to firgure that oug. I even looked in a math book. Nothing.
T
To find the height of the second pebble when the first pebble strikes the ground, we need to determine the time it takes for the first pebble to fall one foot and then use that time to calculate the height of the second pebble.
We know that when an object is released from rest and falls, its height can be described by the equation h = 16t^2, where h is the height in feet and t is the time in seconds.
Let's start by finding the time it takes for the first pebble to fall one foot.
Given that the height of the first pebble when it falls one foot is h = 1 foot, we can set up the equation as follows:
1 = 16t^2
To solve for t, we divide both sides of the equation by 16:
1/16 = t^2
Taking the square root of both sides, we get:
√(1/16) = √t^2
1/4 = t
So, it takes 1/4 of a second for the first pebble to fall one foot.
Now, we can use this time, t = 1/4 second, to find the height of the second pebble when the first pebble strikes the ground.
Since the height of the cliff is 64 feet, the height of the second pebble would be 64 - 1 = 63 feet (subtracting one foot for the fall of the first pebble).
Therefore, the second pebble is 63 feet above the ground when the first pebble strikes the ground.