A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 16.0 m/s. The cliff is h = 25.0 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?
s
With what speed and angle of impact does the stone land?
m/s
° below the horizontal
0.5g*t^2 = 25 m.
4.9*t^2 = 25, t^2 = 5.1, t = 2.26 s.
Y = Yo + g*t = 0 + 9.8*2.26 = 22.1 m/s. = Ver. component of final velocity.
V = 16 + 22.1i = 27.3m/s[54.1o].
To find the time it takes for the stone to strike the beach below the cliff, we need to use the equations of motion. In this case, since the stone is thrown horizontally, its initial vertical velocity is zero.
First, let's find the time it takes for the stone to fall from the cliff to the beach. We can use the equation:
h = (1/2) * g * t^2
Where h is the height of the cliff (25.0 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Rearranging the equation, we have:
t^2 = (2 * h) / g
t^2 = (2 * 25.0) / 9.8
t^2 = 5.102
t = sqrt(5.102)
t = 2.26 seconds (approx)
So, it takes approximately 2.26 seconds for the stone to fall from the cliff to the beach below.
Next, let's find the horizontal distance the stone travels in this time. Since the initial horizontal velocity is 16.0 m/s and the time is 2.26 seconds, we can use the equation:
d = v * t
Where d is the distance, v is the initial horizontal velocity, and t is the time. Plugging in the values, we get:
d = 16.0 * 2.26
d = 36.16 meters (approx)
Therefore, the stone lands approximately 36.16 meters away from the base of the cliff.
Now, let's find the speed and angle of impact when the stone lands. Since the stone only has an initial horizontal velocity, the vertical velocity remains zero. We can find the magnitude of the final velocity using the equation:
v = sqrt((v_initial_horizontal)^2 + (v_initial_vertical)^2)
Since v_initial_vertical is zero, the equation simplifies to:
v = v_initial_horizontal
So, the speed of impact is 16.0 m/s.
To find the angle of impact below the horizontal, we can use the equation:
θ = tan^(-1)(v_initial_vertical / v_initial_horizontal)
Since the v_initial_vertical is zero, the equation simplifies to:
θ = tan^(-1)(0 / 16.0)
θ = tan^(-1)(0)
θ = 0°
Therefore, the stone lands with a speed of 16.0 m/s and an angle of impact of 0° below the horizontal.