Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.34 m. The stones are thrown with the same speed of 9.35 m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.
To find the location where the stones cross paths, we need to determine the time it takes for each stone to reach that point.
Let's first find the time it takes for the stone thrown upward to reach the crossing point:
We can use the kinematic equation:
y = y0 + v0t + (1/2)at^2
where:
y is the displacement (change in height),
y0 is the initial height,
v0 is the initial velocity,
t is the time, and
a is the acceleration.
For the stone thrown upward:
y0 = 0 m (base of the cliff)
v0 = 9.35 m/s (initial velocity)
a = -9.8 m/s^2 (acceleration due to gravity, negative because it is in the opposite direction of the motion)
Since we are interested in the time it takes for the stone to reach the crossing point, we set y = 6.34 m (height of the cliff):
6.34 m = 0 m + 9.35 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Rearranging the equation:
-4.9 t^2 + 9.35 t + 6.34 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where:
a = -4.9,
b = 9.35, and
c = 6.34.
Solving this equation gives two possible values for t. We select the positive value since time cannot be negative.
Now, let's find the time it takes for the stone thrown downward to reach the crossing point:
Using the same equation, but with a positive acceleration this time (due to gravity pulling the stone downward):
6.34 m = 0 m - 9.35 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Simplifying:
4.9 t^2 - 9.35 t + 6.34 = 0
Again, using the quadratic formula, we solve for t.
Once we have the time values for both stones, we can find the location above the base of the cliff where they cross paths by multiplying the time each stone takes by its initial velocity:
Location = 9.35 m/s * t1 = -9.35 m/s * t2