A rock is dropped off the edge of the cliff and its distance S ( in feet ) from the top of the cliff after T seconds is S ( T ) = 16t^2. Assume the distance from the top ofthe cliff to the water below is 96ft
Make a table of the average velocity and approximate the velocity at which the rock strikes the water?
the rock hits in √6 seconds. My table is
t v (at)
0 0
√6 32√6
avg speed: 32√6/√6 = 32
t s (1/2 at^2)
0 0
√6 96
avg speed: 96/√6 = 16√6
Ya gotta be specific to get what you want.
Thank you that was the exact question. I wish it was more specific
To find the average velocity of the rock and approximate the velocity at which it strikes the water, we can use the formula for average velocity, which is the change in distance divided by the change in time.
Let's create a table to calculate the average velocity at different time intervals and estimate the velocity at which the rock strikes the water.
Time Interval (T) | Distance (S) | Average Velocity (V)
-----------------------------------------------------------------------
0 sec | 0 ft | N/A
1 sec | 16 ft | N/A
2 sec | 64 ft | 32 ft/sec
3 sec | 144 ft | 56 ft/sec
4 sec | 256 ft | 80 ft/sec
5 sec | 400 ft | 104 ft/sec
6 sec | 576 ft | 128 ft/sec
7 sec | 784 ft | 152 ft/sec
8 sec | 1024 ft | 176 ft/sec
To estimate the velocity at which the rock strikes the water, we can look at the last few time intervals. From the table, we can see that as time increases, the average velocity of the rock increases. At 8 seconds, the average velocity is approximately 176 ft/sec.
Therefore, we can estimate that the velocity at which the rock strikes the water is around 176 ft/sec.
To find the average velocity of the rock at different time intervals, we need to calculate the change in distance divided by the change in time. We can use this information to approximate the velocity at which the rock strikes the water, which is when the distance from the top of the cliff to the water is 96 feet.
Let's create a table to calculate the average velocity at different time intervals:
| Time (seconds) | Distance (feet) from the top | Change in Distance (ΔS) | Average Velocity (V) |
|----------------|-----------------------------|------------------------|----------------------|
| 0 | 0 | 0 | - |
| 1 | 16 | 16 | |
| 2 | 64 | 48 | |
| 3 | 144 | 80 | |
| 4 | 256 | 112 | |
| 5 | 400 | 144 | |
To calculate the change in distance (ΔS), we subtract the previous distance from the current distance. For example, at t=1 (time = 1 second), ΔS = S(1) - S(0) = 16 - 0 = 16.
To approximate the average velocity (V), we divide the change in distance (ΔS) by the change in time (ΔT). For example, at t=1 (time = 1 second), V = ΔS / ΔT = 16 / 1 = 16 feet per second.
Now, let's fill in the table with the calculations:
| Time (seconds) | Distance (feet) from the top | Change in Distance (ΔS) | Average Velocity (V) |
|----------------|-----------------------------|------------------------|----------------------|
| 0 | 0 | 0 | - |
| 1 | 16 | 16 | 16 |
| 2 | 64 | 48 | 48 |
| 3 | 144 | 80 | 80 |
| 4 | 256 | 112 | 112 |
| 5 | 400 | 144 | 144 |
To approximate the velocity at which the rock strikes the water, we need to find the time when the distance from the top of the cliff to the water is 96 feet. Looking at the table, we can see that when the time is around 4 seconds, the distance is closest to 96 feet.
Since our measurements are discrete (based on seconds), we can estimate the velocity at 4 seconds by taking the average velocity between the time intervals 3-4 seconds and 4-5 seconds.
The average velocity at 4 seconds is:
V(4) ≈ (V(3) + V(5)) / 2
Substituting the values from the table:
V(4) ≈ (112 + 144) / 2 = 128 feet per second
Therefore, the approximate velocity at which the rock strikes the water is 128 feet per second.