At t = 0, a stone is dropped from a cliff above a lake; 2.1 seconds later another stone is thrown downward from the same point with an initial speed of 33 m/s. Both stones hit the water at the same instant. Find the height of the cliff
To find the height of the cliff, we need to calculate the time it takes for the first stone to hit the water.
Let's assume the height of the cliff is H meters.
For the first stone:
We know that the acceleration due to gravity is approximately 9.8 m/s^2 (ignoring air resistance). The initial velocity of the stone is 0 m/s (dropped from rest). The final velocity when it hits the water will also be 0 m/s (it comes to a stop). Using these values, we can use the kinematic equation for vertical motion:
Final Velocity = Initial Velocity + Acceleration * Time
0 m/s = 0 m/s + (9.8 m/s^2) * Time1
This equation simplifies to:
9.8 * Time1 = 0
Time1 = 0 seconds
We find that the first stone takes 0 seconds to hit the water.
Now, let's focus on the second stone thrown downwards.
The initial velocity of the second stone is 33 m/s. The time it takes for the second stone to hit the water is given as 2.1 seconds after the first stone was dropped. We can again use the same kinematic equation:
Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2
Plugging in the values we know:
H = (33 m/s) * Time2 + (1/2) * (9.8 m/s^2) * (Time2)^2
Since we know that Time2 = Time1 + 2.1 seconds, we can substitute this value in the equation:
H = (33 m/s) * (Time1 + 2.1 s) + (1/2) * (9.8 m/s^2) * ((Time1 + 2.1 s)^2)
Since we already calculated Time1 to be 0 seconds, the equation further simplifies to:
H = (33 m/s) * (0 s + 2.1 s) + (1/2) * (9.8 m/s^2) * ((0 s + 2.1 s)^2)
Simplifying further:
H = (33 m/s) * (2.1 s) + (1/2) * (9.8 m/s^2) * (2.1 s)^2
H = 69.3 m + 1/2 * 9.8 m/s^2 * 4.41 s^2
H = 69.3 m + 21.5049 m
H = 90.8049 m
Therefore, the height of the cliff is approximately 90.8 meters.