A stone is thrown horizontally at 30.0m/s from the top of a very tall cliff. Calculate its horizontal position and vertical position at 2s intervals for the first 10.0 s.
To calculate the horizontal and vertical positions of the stone at different time intervals, we need to use the equations of motion.
First, let's start by calculating the horizontal position. Since the stone is thrown horizontally, its horizontal velocity remains constant throughout its motion. The horizontal position of the stone can be calculated using the equation:
horizontal position = initial horizontal velocity * time
Given that the stone is thrown horizontally at a velocity of 30.0 m/s and we want to calculate the horizontal position at 2-second intervals for the first 10.0 seconds, we can set up a table to calculate the values:
Time (s) | Horizontal position (m)
------------------------------------------
0 (initial) | 0
2 | 30.0 * 2 = 60.0
4 | 30.0 * 4 = 120.0
6 | 30.0 * 6 = 180.0
8 | 30.0 * 8 = 240.0
10 | 30.0 * 10 = 300.0
Therefore, the horizontal position of the stone at 2-second intervals for the first 10.0 seconds is 0m, 60m, 120m, 180m, 240m, and 300m.
Next, let's calculate the vertical position of the stone at different time intervals. Considering that the stone is thrown from the top of a cliff, we need to take into account gravity's effect on its vertical motion. The equation to calculate the vertical position of the stone is:
vertical position = initial vertical velocity × time + 0.5 × acceleration due to gravity × time^2
The initial vertical velocity is 0 m/s since the stone is only thrown horizontally. The acceleration due to gravity is approximately 9.8 m/s^2.
Again, using a table to calculate the values:
Time (s) | Vertical position (m)
---------------------------------------------------
0 (initial) | 0
2 | 0 + 0.5 * 9.8 * (2)^2 = 19.6
4 | 0 + 0.5 * 9.8 * (4)^2 = 78.4
6 | 0 + 0.5 * 9.8 * (6)^2 = 176.4
8 | 0 + 0.5 * 9.8 * (8)^2 = 313.6
10 | 0 + 0.5 * 9.8 * (10)^2 = 490.0
Therefore, the vertical position of the stone at 2-second intervals for the first 10.0 seconds is 0m, 19.6m, 78.4m, 176.4m, 313.6m, and 490.0m.
Note: The calculations assume no air resistance and a constant gravitational acceleration of 9.8 m/s^2.
come on. This is very straightforward.
horizontal velocity stays constant, 30m/s, so hor. pos. = 30*t
distance fallen = 1/2 at² = 4.9t²