A Student stands at the edge of a cliff and throws a stone horizontally over the
edge with a speed of 18.0 m/s The cliff is 50 m above a flat horizontal beach. (a)
How long after being released does the stone strike the beach below the cliff?
(b) With what speed and angle of impact does it land?
First we need to find the time it takes for the stone to hit the beach. We can use the equation:
y = v_iy * t + 0.5 * a_y * t^2
where
y = -50 (height of the cliff)
v_iy = 0 (initial vertical velocity - stone is thrown horizontally)
a_y = -9.8 m/s^2 (acceleration due to gravity)
Plugging in the values, we get:
-50 = 0 + 0.5 * (-9.8) * t^2
-50 = -4.9t^2
t^2 = 10.2
t = sqrt(10.2)
t = 3.19 s
So it takes 3.19 seconds for the stone to hit the beach.
Next, we need to find the horizontal distance the stone travels. We can use the equation:
x = v_ix * t
where
v_ix = 18.0 m/s (initial horizontal velocity)
Plugging in the values, we get:
x = 18.0 * 3.19
x = 57.42 m
So the stone lands 57.42 meters away horizontally from where it was thrown.
Finally, we can find the speed and angle of impact. We first need to find the vertical velocity when the stone hits the beach:
v_fy = v_iy + a_y * t
v_fy = 0 + (-9.8) * 3.19
v_fy = -31.24 m/s
Now we can find the speed of impact:
v_impact = sqrt(v_ix^2 + v_fy^2)
v_impact = sqrt((0)^2 + (-31.24)^2)
v_impact = 31.24 m/s
The angle of impact can be found using:
θ = arctan(v_fy / v_ix)
θ = arctan(-31.24 / 18.0)
θ = -59.03 degrees
Therefore, the stone lands on the beach with a speed of 31.24 m/s and at an angle of -59.03 degrees.