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:

Step1: Identify knowns and unknowns

We know the following:

• The time when the rock passes the thrower on the way down: \( t_1 = 5.00 \, \text{s} \)
• The vertical displacement from the thrower's hand to the water: \( y = - 24.0 \, \text{m} \) (negative because it's downward)
• Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \)

We need to find the initial velocity \( v_0 \) first, then find the velocity when it hits the water.

Step2: Use the kinematic equation for displacement

The kinematic equation for vertical displacement is:
\[ y = v_0 t + \frac{1}{2} a t^2 \]
When the rock passes the thrower on the way down at \( t = 5.00 \, \text{s} \), the displacement from the initial position is \( y = 0 \) (since it's back to the starting height). So we can solve for \( v_0 \):
\[ 0 = v_0 (5.00) + \frac{1}{2} (-9.8)(5.00)^2 \]
\[ 0 = 5v_0 - 122.5 \]
\[ 5v_0 = 122.5 \]
\[ v_0 = \frac{122.5}{5} = 24.5 \, \text{m/s} \]

Step3: Use the kinematic equation for velocity when hitting water

Now we use the kinematic equation \( v = v_0 + a t \) to find the velocity when it hits the water. First, we need to find the time \( t \) when the rock hits the water. We use the displacement equation again with \( y = -24.0 \, \text{m} \):
\[ -24.0 = 24.5 t + \frac{1}{2} (-9.8) t^2 \]
\[ -24.0 = 24.5 t - 4.9 t^2 \]
\[ 4.9 t^2 - 24.5 t - 24.0 = 0 \]
This is a quadratic equation \( at^2 + bt + c = 0 \) with \( a = 4.9 \), \( b = -24.5 \), \( c = -24.0 \). Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{24.5 \pm \sqrt{(-24.5)^2 - 4(4.9)(-24.0)}}{2(4.9)} \]
\[ t = \frac{24.5 \pm \sqrt{600.25 + 470.4}}{9.8} \]
\[ t = \frac{24.5 \pm \sqrt{1070.65}}{9.8} \]
\[ t = \frac{24.5 \pm 32.72}{9.8} \]
We take the positive root (since time can't be negative):
\[ t = \frac{24.5 + 32.72}{9.8} \approx \frac{57.22}{9.8} \approx 5.84 \, \text{s} \]

Now we find the velocity at \( t = 5.84 \, \text{s} \):
\[ v = v_0 + a t \]
\[ v = 24.5 + (-9.8)(5.84) \]
\[ v = 24.5 - 57.23 \]
\[ v \approx -32.73 \, \text{m/s} \]
The negative sign indicates the velocity is downward. The magnitude is approximately \( 32.7 \, \text{m/s} \).

(Alternatively, we can use the equation \( v^2 = v_0^2 + 2 a y \) directly to find the velocity when hitting the water. Let's check that method for accuracy.)

Using \( v^2 = v_0^2 + 2 a y \):
\[ v^2 = (24.5)^2 + 2(-9.8)(-24.0) \]
\[ v^2 = 600.25 + 470.4 \]
\[ v^2 = 1070.65 \]
\[ v = \pm \sqrt{1070.65} \approx \pm 32.72 \, \text{m/s} \]
Since the rock is moving downward when it hits the water, the velocity is \( -32.7 \, \text{m/s} \) (or \( 32.7 \, \text{m/s} \) downward).

Answer:

The speed of the rock just before it reaches the water is approximately \(\boxed{32.7}\) m/s (the velocity is \(-32.7\) m/s, indicating downward direction, but speed is the magnitude, so \(32.7\) m/s).