a rock is tossed straight up with a speed of 30m/s. when it returns it falls into a hole 10m deep. what is the rocks velocity as it hits the bottom of the whole? how long would the rock in the air, form the instant it is released until it hits the bottom on the hole
s = vot + 1/2 at^2 = 30t - 4.9t^2
solve for s = -10. t = 6.439s
v = 30 - 9.8*6.439 = -33.1m/s
Makes sense, since it will be going -30m/s as it starts into the hole.
To find the rock's velocity as it hits the bottom of the hole, we need to consider the two phases of its motion: when it is moving upward and when it is moving downward.
Step 1: Find the time it takes for the rock to reach its highest point (when it momentarily stops moving upward):
The initial velocity (v0) is 30 m/s.
The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Using the equation:
v = v0 + at
0 = 30 - 9.8t
9.8t = 30
t = 30 / 9.8
t ≈ 3.06 seconds (rounded to two decimal places)
So it takes approximately 3.06 seconds for the rock to reach its highest point.
Step 2: Find the time it takes for the rock to fall from its highest point to the bottom of the hole:
The height of the hole (h) is 10 m.
Using the equation for gravitational potential energy:
PE = mgh
mgh = mv^2/2 (since v = 0 at the highest point)
gh = v^2/2
v^2 = 2gh
v = √(2gh)
v = √(2 * 9.8 * 10)
v ≈ 14 m/s (rounded to two decimal places)
The velocity as the rock hits the bottom of the hole is approximately 14 m/s.
Step 3: Find the total time the rock is in the air:
The time it takes for the rock to reach its highest point is 3.06 seconds (from Step 1).
The time it takes for the rock to fall from its highest point to the bottom of the hole is the same as the time it takes to reach the highest point (symmetrical motion).
So, the total time the rock is in the air is approximately 3.06 seconds.
To find the velocity of the rock as it hits the bottom of the hole, we need to consider the motion of the rock both while it is moving upward and when it is falling downward.
First, let's calculate the time it takes for the rock to reach its maximum height. We can use the equation:
v = u + at
where:
v = final velocity (0 m/s, since the rock comes to a stop at the top)
u = initial velocity (30 m/s)
a = acceleration (due to gravity, approximately -9.8 m/s^2, since it acts in the opposite direction to the motion)
t = time
Rearranging the equation, we have:
t = (v - u) / a
Substituting the values, we find:
t = (0 m/s - 30 m/s) / -9.8 m/s^2
t ≈ 3.06 seconds
So, it takes approximately 3.06 seconds for the rock to reach its maximum height.
Next, let's calculate the velocity of the rock as it falls into the hole. Since the rock falls from rest at the top, we can use the equation:
s = ut + (1/2)at^2
where:
s = displacement (the depth of the hole, 10 m)
u = initial velocity (0 m/s)
a = acceleration (due to gravity, approximately -9.8 m/s^2)
t = time
Rearranging the equation, we have:
s = (1/2)at^2
Substituting the values, we find:
10 m = (1/2)(-9.8 m/s^2)(3.06 s)^2
10 m ≈ -45 m
Since the depth of the hole is positive, the negative sign indicates that it's in the opposite direction. Therefore, the rock has a velocity of approximately 45 m/s downwards when it hits the bottom of the hole.
Additionally, we can calculate the total time the rock is in the air. This can be done by adding the time it takes to reach the maximum height (3.06 seconds) to the time it takes to fall back to the ground. Since the rocks fall time is equal to the time it takes to reach the maximum height, the total time in the air would be approximately 6.12 seconds.