A stone is thrown horizontally at 30.0m/s {\rm m}/{\rm s} from the top of a very tall cliff. Calculate its horizontal position and vertical position at 2 s {\rm s} intervals for the first 10.0s {\rm s}
a)Calculate the horizontal x and vertical y positions at time t = 2 s
To calculate the horizontal position (x) and vertical position (y) of a stone at time t = 2s, we first need to determine the initial horizontal and vertical velocities.
Given:
Initial horizontal velocity (Vx) = 30.0 m/s (thrown horizontally)
Initial vertical velocity (Vy) = 0 m/s (thrown horizontally)
Horizontal position (x) at time t = 2s:
Since the stone is thrown horizontally, its horizontal velocity remains constant throughout its motion. Therefore, the horizontal distance covered by the stone at any time can be calculated using the formula: x = Vx * t.
Substituting the given values:
x = 30.0 m/s * 2s
x = 60.0 m
Therefore, the horizontal position (x) of the stone at t = 2s is 60.0 meters.
Vertical position (y) at time t = 2s:
To calculate the vertical position, we need to consider the effect of gravity. The vertical distance covered by the stone can be calculated using the formula: y = Vy * t + (1/2) * g * t^2.
Since the stone is thrown horizontally, the initial vertical velocity (Vy) is 0 m/s. Also, the acceleration due to gravity (g) is approximately 9.8 m/s^2.
Substituting the given values:
y = 0 m/s * 2s + (1/2) * 9.8 m/s^2 * (2s)^2
y = (1/2) * 9.8 m/s^2 * 4s^2
y = 19.6 m/s^2 * 4s^2
y = 78.4 m
Therefore, the vertical position (y) of the stone at t = 2s is 78.4 meters.