A rock is thrown horizontally with a speed of 5.9 m/s from a height of 63.5 meters from the ground below. With what vertical speed does the rock hit the ground?
To find the vertical speed with which the rock hits the ground, we can use the equation of motion. The equation of motion for vertical motion is:
h = u*t + (1/2)*g*t^2
where h represents the height, u represents the initial velocity, t represents time, and g represents the acceleration due to gravity (9.8 m/s^2).
In this case, the rock is thrown horizontally, so the initial vertical velocity (u) is 0. Therefore, we can remove the u*t term from the equation. The equation becomes:
h = (1/2)*g*t^2
We can rearrange the equation to solve for t:
t = sqrt(2h/g)
Substituting the values given:
h = 63.5 m
g = 9.8 m/s^2
t = sqrt(2 * 63.5 / 9.8)
= sqrt(127 / 9.8)
≈ 4.04 s
Now that we have the time it takes for the rock to hit the ground, we can find the vertical speed (v) using the equation:
v = g*t
Substituting the values:
v = 9.8 m/s^2 * 4.04 s
≈ 39.6 m/s
Therefore, the rock hits the ground with a vertical speed of approximately 39.6 m/s.