Two diamonds begin a free fall from rest from the same height 0.9 s apart. How long after the first object begins to fall will the two objects be 11 m apart?
Let t=0 be the time the first one is dropped. The distance that it falls after that is
y1 = (g/2)t^2 = 4.9 t^2
The distance that the second one has fallen at time t(>0.9s) is
y2 = (g/2)(t-0.9)^2 = 4.9(t-0.9)^2
You want to solve for the time t when
y1 - y2 = 4.9 [t^2 - (t-1)^2] = 11 m
2t - 1 = 11/4.9 = 2.24 s
t = 1.62 s
Two diamonds begin a free fall from rest from the same height 1.2 s apart. How long after the first object begins to fall will the two objects be 20 m apart?
To solve this problem, we need to consider the motion of both diamonds and find the time at which they will be 11 meters apart.
First, let's define some variables:
- Let t1 be the time at which the first diamond starts to fall.
- Let t2 be the time at which the second diamond starts to fall.
We know that the second diamond starts falling 0.9 seconds after the first diamond, so we can write:
t2 = t1 + 0.9 seconds
Now, let's consider the motion of each diamond. Assuming there is no air resistance, both diamonds will fall with the same acceleration (acceleration due to gravity), and the distance traveled by an object in free fall can be calculated using the equation:
d = (1/2) * g * t^2
where d is the distance traveled, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
For the first diamond, the distance traveled can be expressed as:
d1 = (1/2) * g * (t - t1)^2
For the second diamond, the distance traveled can be expressed as:
d2 = (1/2) * g * (t - t2)^2
We want the two diamonds to be 11 meters apart, so we have:
d1 - d2 = 11 meters
Substituting the expressions for d1 and d2, we get:
(1/2) * g * (t - t1)^2 - (1/2) * g * (t - t2)^2 = 11
Simplifying the equation, we have:
g * ((t - t1)^2 - (t - t2)^2) = 22
Now, we can substitute the value of t2 in terms of t1 to simplify the equation further:
g * ((t - t1)^2 - (t - (t1 + 0.9))^2) = 22
After simplification, we get a quadratic equation in terms of t1:
0.81g * t1^2 - 2.43g * t1 + 22 = 0
Solving this quadratic equation will give us the value of t1 at which the two objects will be 11 meters apart.