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 6.02 m. The stones are thrown with the same speed of 9.20 m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.
To find the point where the two stones cross paths, we need to determine the time it takes for each stone to reach that point.
Let's start by considering the stone thrown upward. We can use the kinematic equation for vertical motion:
Δy = v_0t - (1/2)gt^2
Where:
Δy = Final position (height above the base of the cliff)
v_0 = Initial velocity (9.20 m/s)
t = Time taken
g = Acceleration due to gravity (-9.8 m/s^2)
Since the stone is thrown upward, the Δy will be positive and equal to the height of the cliff (6.02 m). We can plug in these values and solve for t:
6.02 = 9.20t - (1/2)(9.8)t^2
Simplifying the equation, we get a quadratic equation:
4.9t^2 - 9.20t + 6.02 = 0
Now let's consider the stone thrown downward. The initial velocity and acceleration due to gravity remain the same. This time, the Δy will be negative since the stone is thrown downward. We want to find the time it takes for this stone to reach the same point above the base of the cliff. We can set up the equation as:
-6.02 = -9t - (1/2)(9.8)t^2
Simplifying, we get another quadratic equation:
4.9t^2 + 9t + 6.02 = 0
Now we need to solve these two quadratic equations simultaneously to find the common value of t (time) when the stones cross paths.
Once we find the value of t, we can substitute it back into either of the two equations to find the height above the base of the cliff where the stones cross paths.