A stone is thrown horizontally at 30.0 {\rm m}/{\rm s} from the top of a very tall cliff. Calculate its horizontal position and vertical position at 2 {\rm s} intervals for the first 10.0 {\rm s}.
To calculate the horizontal and vertical positions of the stone at 2-second intervals for the first 10 seconds, we need to use the equations of motion.
Let's start by calculating the horizontal position at each interval.
1. Initial horizontal position (t = 0s): The stone is thrown horizontally, so its initial horizontal position is 0 meters.
2. Horizontal velocity (v_x): The stone is thrown horizontally at 30.0 m/s, so its horizontal velocity remains constant throughout its motion. Therefore, the horizontal velocity, v_x, is 30.0 m/s.
3. Horizontal displacement (x): The horizontal displacement can be calculated using the equation: x = v_x * t.
Let's calculate the horizontal positions at each interval:
t = 2s: x = v_x * t = 30.0 m/s * 2s = 60.0 m
t = 4s: x = 30.0 m/s * 4s = 120.0 m
t = 6s: x = 30.0 m/s * 6s = 180.0 m
t = 8s: x = 30.0 m/s * 8s = 240.0 m
t = 10s: x = 30.0 m/s * 10s = 300.0 m
So, the horizontal positions of the stone at 2-second intervals for the first 10.0 seconds are:
At t = 0s, x = 0m
At t = 2s, x = 60.0m
At t = 4s, x = 120.0m
At t = 6s, x = 180.0m
At t = 8s, x = 240.0m
At t = 10s, x = 300.0m
Now, let's calculate the vertical positions at each interval.
1. Initial vertical position (t = 0s): The stone is thrown horizontally, so its initial vertical position is the same as the height of the cliff.
2. Acceleration (a_y): The stone is only affected by the acceleration due to gravity, which is approximately 9.8 m/s^2 downwards. Therefore, the acceleration in the vertical direction, a_y, is -9.8 m/s^2.
3. Vertical displacement (y): The vertical displacement can be calculated using the equation: y = v_{y_0} * t + (1/2) * a_y * t^2, where v_{y_0} is the initial vertical velocity.
At t = 0s, y = 0m (initial height of the cliff)
t = 2s: y = v_{y_0} * t + (1/2) * a_y * t^2 = 0m + (1/2) * (-9.8 m/s^2) * (2s)^2 = -19.6 m
t = 4s: y = v_{y_0} * t + (1/2) * a_y * t^2 = 0m + (1/2) * (-9.8 m/s^2) * (4s)^2 = -78.4 m
t = 6s: y = v_{y_0} * t + (1/2) * a_y * t^2 = 0m + (1/2) * (-9.8 m/s^2) * (6s)^2 = -176.4 m
t = 8s: y = v_{y_0} * t + (1/2) * a_y * t^2 = 0m + (1/2) * (-9.8 m/s^2) * (8s)^2 = -313.6 m
t = 10s: y = v_{y_0} * t + (1/2) * a_y * t^2 = 0m + (1/2) * (-9.8 m/s^2) * (10s)^2 = -490.0 m
Since the stone is thrown horizontally, it does not travel vertically. The negative values indicate that the stone has fallen below the initial height of the cliff.
So, the vertical positions of the stone at 2-second intervals for the first 10.0 seconds are:
At t = 0s, y = 0m (initial height of the cliff)
At t = 2s, y = -19.6m
At t = 4s, y = -78.4m
At t = 6s, y = -176.4m
At t = 8s, y = -313.6m
At t = 10s, y = -490.0m
To calculate the horizontal and vertical positions of the stone at various time intervals, we can make use of the kinematic equations of motion.
Let's break down the given information. The initial horizontal velocity of the stone, v_horizontal, is 30.0 m/s. Since the stone is thrown horizontally, there is no initial vertical velocity component. The only force acting on the stone is gravity, pulling it down vertically.
To find the horizontal position at each time interval, we can use the equation:
d_horizontal = v_horizontal * t
where d_horizontal is the horizontal position, v_horizontal is the initial horizontal velocity, and t is the time elapsed.
For the vertical position, we need to consider the effect of gravity. The equation for the vertical position at any given time, t, is:
d_vertical = v_vertical_initial * t + (1/2) * a * t^2
where d_vertical is the vertical position, v_vertical_initial is the initial vertical velocity (in this case, 0 m/s), t is the time elapsed, and a is the acceleration due to gravity (-9.8 m/s^2).
Let's calculate the horizontal and vertical positions at 2-second intervals for the first 10 seconds:
At t = 0 s:
Horizontal position: d_horizontal = v_horizontal * t = 30.0 m/s * 0 s = 0 m
Vertical position: d_vertical = v_vertical_initial * t + (1/2) * a * t^2 = 0 m/s * 0 s + (1/2) * (-9.8 m/s^2) * (0 s)^2 = 0 m
At t = 2 s:
Horizontal position: d_horizontal = v_horizontal * t = 30.0 m/s * 2 s = 60.0 m
Vertical position: d_vertical = v_vertical_initial * t + (1/2) * a * t^2 = 0 m/s * 2 s + (1/2) * (-9.8 m/s^2) * (2 s)^2 = -19.6 m
At t = 4 s:
Horizontal position: d_horizontal = v_horizontal * t = 30.0 m/s * 4 s = 120.0 m
Vertical position: d_vertical = v_vertical_initial * t + (1/2) * a * t^2 = 0 m/s * 4 s + (1/2) * (-9.8 m/s^2) * (4 s)^2 = -78.4 m
At t = 6 s:
Horizontal position: d_horizontal = v_horizontal * t = 30.0 m/s * 6 s = 180.0 m
Vertical position: d_vertical = v_vertical_initial * t + (1/2) * a * t^2 = 0 m/s * 6 s + (1/2) * (-9.8 m/s^2) * (6 s)^2 = -176.4 m
At t = 8 s:
Horizontal position: d_horizontal = v_horizontal * t = 30.0 m/s * 8 s = 240.0 m
Vertical position: d_vertical = v_vertical_initial * t + (1/2) * a * t^2 = 0 m/s * 8 s + (1/2) * (-9.8 m/s^2) * (8 s)^2 = -313.6 m
At t = 10 s:
Horizontal position: d_horizontal = v_horizontal * t = 30.0 m/s * 10 s = 300.0 m
Vertical position: d_vertical = v_vertical_initial * t + (1/2) * a * t^2 = 0 m/s * 10 s + (1/2) * (-9.8 m/s^2) * (10 s)^2 = -490.0 m
Therefore, the horizontal and vertical positions at 2-second intervals for the first 10 seconds are:
Time (s) Horizontal position (m) Vertical position (m)
--------------------------------------------------------
0 0 0
2 60.0 -19.6
4 120.0 -78.4
6 180.0 -176.4
8 240.0 -313.6
10 300.0 -490.0