You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 7.00 s after it was thrown. What is the speed of the rock just before it reaches the water 29.0 m below the point where the rock left your hand? Ignore air resistance.
7.4
So the issue is what speed did you throw it with?
it takes 3.4 seconds to get to the apex...at the top it has zero speed.
vf=vi+gt
0=vi-9.8*3.5
solve for the vi initiual velocity.
then
final KE =initial KE +initial PE
1/2 m vf^2=1/2 m vi^2 + mg*28
solve for vf
To find the speed of the rock just before it reaches the water, we can use the equations of motion for vertical motion in free fall.
Let's break down the problem step-by-step:
Step 1: Identify known quantities:
- Initial velocity (u) = unknown (since the problem doesn't mention it, we assume it's thrown with an initial velocity of 0 m/s)
- Final velocity (v) = ? (to be determined)
- Time (t) = 7.00 s
- Acceleration due to gravity (g) = -9.8 m/s² (taking negative because it acts in the opposite direction of the positive direction)
Step 2: Determine the distance traveled by the rock during the upward motion:
The total time of flight of the rock is 7.00 s. Since the rock passes you on its way down, we can assume that the upward motion took half of the total time.
The time taken during the upward motion (t_up) = 7.00 s / 2 = 3.50 s
During the upward motion, the rock covers a distance given by the formula:
d = ut + (1/2)gt²
Substituting the values:
0 = (0)(3.50) + (1/2)(-9.8)(3.50)²
Step 3: Solve the equation:
0 = 0 + (-4.9)(3.50)²
0 = -4.9(12.25)
0 = -60.025
Step 4: Determine the distance traveled by the rock during the downward motion:
Now, we can calculate the distance traveled by the rock during the downward motion. This distance is equal to the height of the bridge, which is given as 29.0 m.
Step 5: Find the final velocity of the rock:
We can use the formula for final velocity in free fall motion:
v = u + gt
Substituting the values:
v = 0 + (-9.8)(7.00)
Step 6: Solve the equation:
v = -9.8(7.00)
v = -68.6 m/s
Step 7: Calculate the speed:
The speed is the magnitude of the velocity, so we take the absolute value:
Speed = |v| = |-68.6| = 68.6 m/s
Therefore, the speed of the rock just before it reaches the water is 68.6 m/s.
To find the speed of the rock just before it reaches the water, we can use the kinematic equation for motion in the vertical direction.
The equation we'll use is:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (the speed of the rock just before it reaches the water)
vi = initial velocity (the speed at which the rock was thrown upward)
a = acceleration due to gravity (which is -9.8 m/s^2 since it's downward)
d = displacement (which is 29.0 m in this case)
First, let's find the initial velocity of the rock when it was thrown upward. We know that it passed you on its way down 7.00 seconds after it was thrown.
Since the rock is thrown upward and then comes back down, we can split the motion into two parts: upward and downward.
In the upward part, the time taken to reach maximum height is the same as the time taken to come back down to the starting point. Therefore, the time taken for the upward motion is half of the total time, which is 7.00 seconds divided by 2, which equals 3.50 seconds.
Now, we can calculate the initial velocity using the equation:
vf = vi + at
At maximum height, the final velocity is 0 m/s since the rock momentarily stops before falling back down. Thus, we can rewrite the equation as:
0 = vi - 9.8 * 3.50
Solving for vi:
vi = 9.8 * 3.50
vi ≈ 34.30 m/s
Now that we have the initial velocity, we can calculate the final velocity using the equation mentioned earlier:
vf^2 = vi^2 + 2ad
Substituting the known values:
vf^2 = (34.30)^2 + 2 * (-9.8) * 29.0
vf^2 ≈ 1175.49 + (-568.4)
vf^2 ≈ 607.09
Taking the square root of both sides:
vf ≈ √607.09
vf ≈ 24.65 m/s
Therefore, the speed of the rock just before it reaches the water is approximately 24.65 m/s.