Question

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, 5.00 s after it was thrown. what is the speed of the rock just before it reaches the water 24.0 m below the point where the rock left your hand? ignore air resistance. express your answer with the appropriate units.

Explanation:

To solve this problem, we can use the kinematic equations of motion under gravity. Let's assume the initial velocity of the rock when thrown upward is \( u \), the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) (acting downward), and we need to find the velocity \( v \) when the rock reaches a point \( 24.0 \, \text{m} \) below the initial point after \( t = 5.00 \, \text{s} \).

Step 1: Identify the known and unknown quantities

• Displacement, \( s = -24.0 \, \text{m} \) (negative because it's downward from the initial position)
• Time, \( t = 5.00 \, \text{s} \)
• Acceleration, \( a = -g = -9.8 \, \text{m/s}^2 \) (negative because it's downward)
• Initial velocity, \( u \) (unknown, but we can find it first or use another equation)
• Final velocity, \( v \) (unknown, what we need to find)

Step 2: Use the second kinematic equation to find the initial velocity \( u \)

The second kinematic equation is:
\[
s = ut + \frac{1}{2}at^2
\]
Substituting the known values:
\[
-24.0 = u(5.00) + \frac{1}{2}(-9.8)(5.00)^2
\]
First, calculate \( \frac{1}{2}(-9.8)(5.00)^2 \):
\[
\frac{1}{2}(-9.8)(25.0) = -4.9 \times 25.0 = -122.5
\]
So the equation becomes:
\[
-24.0 = 5.00u - 122.5
\]
Add \( 122.5 \) to both sides:
\[
5.00u = -24.0 + 122.5 = 98.5
\]
Divide both sides by \( 5.00 \):
\[
u = \frac{98.5}{5.00} = 19.7 \, \text{m/s}
\]

Step 3: Use the first kinematic equation to find the final velocity \( v \)

The first kinematic equation is:
\[
v = u + at
\]
Substituting the known values (\( u = 19.7 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \), \( t = 5.00 \, \text{s} \)):
\[
v = 19.7 + (-9.8)(5.00)
\]
Calculate \( (-9.8)(5.00) \):
\[
-9.8 \times 5.00 = -49.0
\]
So:
\[
v = 19.7 - 49.0 = -29.3 \, \text{m/s}
\]
The negative sign indicates the velocity is downward, which makes sense. The magnitude of the velocity is \( 29.3 \, \text{m/s} \).

Final Answer

The speed of the rock just before it reaches the water is \(\boxed{29.3 \, \text{m/s}}\) (the negative sign indicates direction, but speed is the magnitude, so we can also say \( 29.3 \, \text{m/s} \) downward).

Answer:

To solve this problem, we can use the kinematic equations of motion under gravity. Let's assume the initial velocity of the rock when thrown upward is \( u \), the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) (acting downward), and we need to find the velocity \( v \) when the rock reaches a point \( 24.0 \, \text{m} \) below the initial point after \( t = 5.00 \, \text{s} \).

Step 1: Identify the known and unknown quantities

• Displacement, \( s = -24.0 \, \text{m} \) (negative because it's downward from the initial position)
• Time, \( t = 5.00 \, \text{s} \)
• Acceleration, \( a = -g = -9.8 \, \text{m/s}^2 \) (negative because it's downward)
• Initial velocity, \( u \) (unknown, but we can find it first or use another equation)
• Final velocity, \( v \) (unknown, what we need to find)

Step 2: Use the second kinematic equation to find the initial velocity \( u \)

The second kinematic equation is:
\[
s = ut + \frac{1}{2}at^2
\]
Substituting the known values:
\[
-24.0 = u(5.00) + \frac{1}{2}(-9.8)(5.00)^2
\]
First, calculate \( \frac{1}{2}(-9.8)(5.00)^2 \):
\[
\frac{1}{2}(-9.8)(25.0) = -4.9 \times 25.0 = -122.5
\]
So the equation becomes:
\[
-24.0 = 5.00u - 122.5
\]
Add \( 122.5 \) to both sides:
\[
5.00u = -24.0 + 122.5 = 98.5
\]
Divide both sides by \( 5.00 \):
\[
u = \frac{98.5}{5.00} = 19.7 \, \text{m/s}
\]

Step 3: Use the first kinematic equation to find the final velocity \( v \)

The first kinematic equation is:
\[
v = u + at
\]
Substituting the known values (\( u = 19.7 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \), \( t = 5.00 \, \text{s} \)):
\[
v = 19.7 + (-9.8)(5.00)
\]
Calculate \( (-9.8)(5.00) \):
\[
-9.8 \times 5.00 = -49.0
\]
So:
\[
v = 19.7 - 49.0 = -29.3 \, \text{m/s}
\]
The negative sign indicates the velocity is downward, which makes sense. The magnitude of the velocity is \( 29.3 \, \text{m/s} \).

Final Answer

The speed of the rock just before it reaches the water is \(\boxed{29.3 \, \text{m/s}}\) (the negative sign indicates direction, but speed is the magnitude, so we can also say \( 29.3 \, \text{m/s} \) downward).