12. After an object is projected vertically upwards from a well with a velocity of 30 ms, 4 seconds later another object from the same well is thrown rapidly into the well. If the object projected first collides with the second object, Sri. Find the distance from the point of collision.

We can solve this problem using the equations of motion under constant acceleration due to gravity, which is \( g = 9.8 \, \text{m/s}^2 \). 

### Step 1: Define Variables
- **Object 1**: Projected vertically upwards with an initial velocity \( u_1 = 30 \, \text{m/s} \).
- **Object 2**: Thrown downward from rest (since it is thrown rapidly into the well) after 4 seconds.

Let the time at which the two objects collide be \( t \) seconds after object 1 is projected. Therefore, object 1 has been moving for \( t \) seconds, and object 2 has been moving for \( t - 4 \) seconds (since it was thrown 4 seconds later).

### Step 2: Equations of Motion
For object 1, the vertical displacement \( y_1 \) after time \( t \) is given by:
\[
y_1 = u_1 t - \frac{1}{2} g t^2
\]
\[
y_1 = 30 t - 4.9 t^2
\]

For object 2, the downward displacement \( y_2 \) after time \( t - 4 \) is given by:
\[
y_2 = u_2 (t - 4) + \frac{1}{2} g (t - 4)^2
\]
Since object 2 is thrown downward with an initial velocity of 0 (\( u_2 = 0 \)):
\[
y_2 = 4.9 (t - 4)^2
\]

### Step 3: Condition for Collision
At the point of collision, the displacement of both objects will be the same:
\[
y_1 = y_2
\]
Thus:
\[
30t - 4.9t^2 = 4.9(t - 4)^2
\]
Simplify and solve for \( t \).

### Step 4: Solve for \( t \)
Expanding the right-hand side:
\[
30t - 4.9t^2 = 4.9(t^2 - 8t + 16)
\]
\[
30t - 4.9t^2 = 4.9t^2 - 39.2t + 78.4
\]
Now, collect like terms:
\[
30t - 4.9t^2 = 4.9t^2 - 39.2t + 78.4
\]
\[
0 = 9.8t^2 - 69.2t + 78.4
\]
This is a quadratic equation:
\[
9.8t^2 - 69.2t + 78.4 = 0
\]
Solve using the quadratic formula:
\[
t = \frac{-(-69.2) \pm \sqrt{(-69.2)^2 - 4(9.8)(78.4)}}{2(9.8)}
\]
\[
t = \frac{69.2 \pm \sqrt{4789.44 - 3068.48}}{19.6}
\]
\[
t = \frac{69.2 \pm \sqrt{1720.96}}{19.6}
\]
\[
t = \frac{69.2 \pm 41.49}{19.6}
\]
Thus, we have two solutions for \( t \):
\[
t = \frac{69.2 + 41.49}{19.6} \quad \text{or} \quad t = \frac{69.2 - 41.49}{19.6}
\]
\[
t = 5.64 \, \text{seconds} \quad \text{or} \quad t = 1.42 \, \text{seconds}
\]
Since the second object was thrown after 4 seconds, \( t = 5.64 \) seconds is the valid solution.

### Step 5: Find the Distance from the Point of Collision
Substitute \( t = 5.64 \) seconds into the equation for \( y_1 \):
\[
y_1 = 30t - 4.9t^2
\]
\[
y_1 = 30(5.64) - 4.9(5.64)^2
\]
\[
y_1 = 169.2 - 4.9(31.8)
\]
\[
y_1 = 169.2 - 155.82
\]
\[
y_1 = 13.38 \, \text{m}
\]

Thus, the point of collision occurs at a distance of **13.38 meters** from the point of projection.

---

Would you like to go over any of the steps in more detail?

Here are 5 related questions you can explore:
1. How does gravity affect the motion of objects projected vertically?
2. What are the kinematic equations for uniformly accelerated motion?
3. How would the solution change if the second object was thrown upwards?
4. How can we find the velocity of each object at the moment of collision?
5. What happens when two objects collide with different masses?

**Tip**: Always break down a problem into variables and equations that represent physical quantities. This makes solving complex problems easier.

After an object is projected vertically upwards from a well with a velocity of 30 m/s, 4 seconds later another object from the same well is thrown rapidly into the well. If the object projected first collides with the second object, find the distance from the point of collision.

Explore how to solve a vertical motion problem involving two objects projected from a well, using kinematic equations and the quadratic formula. The solution includes finding the time of collision and the distance from the point of projection. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation