A rock is thrown vertically upward with a speed of 18.0 m/s
 . Exactly 1.00 s
 later, a ball is thrown up vertically along the same path with a speed of 23.0 m/s
 .At what time will they strike each other?

To determine the time at which the rock and the ball will meet, we can use the equations of motion for each object. Here's how we'll approach the problem:

### Step 1: Define the variables
- Let \( t \) be the time after the rock is thrown (in seconds) when the two objects meet.
- Since the ball is thrown 1 second later, the time for the ball when they meet will be \( t - 1 \) seconds.

### Step 2: Use the kinematic equation for each object
The equation for vertical motion under constant acceleration due to gravity (\( g = 9.8 \, \text{m/s}^2 \)) is:

\[
y = v_0 t - \frac{1}{2} g t^2
\]

Where:
- \( y \) is the height above the starting point,
- \( v_0 \) is the initial velocity,
- \( t \) is the time,
- \( g \) is the acceleration due to gravity.

#### Rock’s equation of motion:
The initial velocity of the rock is \( 18.0 \, \text{m/s} \). Its height after \( t \) seconds is:

\[
y_{\text{rock}} = 18.0 t - \frac{1}{2} \times 9.8 t^2 = 18.0 t - 4.9 t^2
\]

#### Ball’s equation of motion:
The ball is thrown 1 second later, with an initial velocity of \( 23.0 \, \text{m/s} \). Its height after \( t - 1 \) seconds is:

\[
y_{\text{ball}} = 23.0 (t - 1) - \frac{1}{2} \times 9.8 (t - 1)^2
\]
\[
y_{\text{ball}} = 23.0 (t - 1) - 4.9 (t - 1)^2
\]

### Step 3: Set the heights equal to each other
To find when the two objects meet, set \( y_{\text{rock}} = y_{\text{ball}} \):

\[
18.0 t - 4.9 t^2 = 23.0 (t - 1) - 4.9 (t - 1)^2
\]

### Step 4: Solve for \( t \)
Let's simplify this equation step by step to find \( t \).

1. Expand both sides.
2. Combine like terms.
3. Solve the resulting quadratic equation.

Let me work that out for you.The time at which the rock and the ball will meet is approximately \( t = 1.89 \, \text{seconds} \) after the rock is thrown.

Since the ball is thrown 1 second later, the ball would have been in motion for \( 1.89 - 1 = 0.89 \, \text{seconds} \) when they meet.

Would you like any further details or have any questions?

Here are 5 related questions to extend your understanding:
1. What is the maximum height reached by each object?
2. How long does it take for each object to reach the ground?
3. What is the relative velocity of the two objects when they meet?
4. How would the meeting time change if the ball was thrown with a different speed?
5. How does air resistance affect the time and position of their meeting?

**Tip:** The symmetry of vertical motion allows you to calculate the total time of flight by doubling the time it takes for an object to reach its highest point.

A rock is thrown vertically upward with a speed of 18.0 m/s. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 23.0 m/s. At what time will they strike each other?

Learn how to solve a vertical motion problem involving a rock and a ball thrown upwards at different times with different speeds. This problem uses kinematic equations and the quadratic formula to find when the two objects meet. Suitable for students in Grades 9-12.

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