A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 22.0 m/s. The cliff is h = 23.0 m above a flat horizontal beach, as shown in the figure below.
How long after being released does the stone strike the beach below the cliff?
With what speed and angle of impact does the stone land?
To solve this problem, we will use the kinematic equations to find the time it takes for the stone to hit the beach, as well as the horizontal distance it travels.
Let's break down the given information:
- Initial speed of the stone, v0 = 22.0 m/s
- Height of the cliff, h = 23.0 m
Since the stone is thrown horizontally, its initial vertical velocity (vy) is 0 m/s. Therefore, its initial velocity vector is purely horizontal.
Using the kinematic equation for vertical motion:
h = v0y * t + (1/2) * a * t^2
Since the stone is in free fall and the only force acting on it is gravity, we know the acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward).
Plugging in the known values:
23.0 m = 0 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation:
23.0 m = (-4.9 m/s^2) * t^2
Rearranging the equation:
4.9 m/s^2 * t^2 = 23.0 m
Dividing both sides by 4.9 m/s^2:
t^2 = 23.0 m / 4.9 m/s^2
Taking the square root of both sides:
t ≈ sqrt(23.0 m / 4.9 m/s^2)
t ≈ 1.37 s
Therefore, it takes approximately 1.37 seconds for the stone to hit the beach.
Now, to find the horizontal distance (x) the stone travels, we need to use the equation for horizontal motion:
x = v0x * t
Since there are no horizontal forces acting on the stone after it is thrown, its horizontal velocity (v0x) remains constant at 22.0 m/s.
Plugging in the known values:
x = 22.0 m/s * 1.37 s
x ≈ 30.1 m
Therefore, the stone travels approximately 30.1 meters horizontally before hitting the beach.