A ball is dropped from the top of an 80 m high tower After 2 s another ball is thrown downwards from the tower. Both the balls reach the ground simultaneously. The initial speed of the second ball is

Options

A. `10 m//s`
B. `20 m//s`
C. `30 m//s`
D. `40 m//s`

Explanation

To solve the problem, we need to analyze the motion of both balls and use the equations of motion under gravity.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Height of the tower (h) = 80 m
- Acceleration due to gravity (g) = 9.8 m/s² (approximately 10 m/s² for simplicity)
- Time taken by the first ball to reach the ground = t seconds
- Time taken by the second ball to reach the ground = t - 2 seconds (since it is thrown 2 seconds later)

2. **Calculate the time taken by the first ball:**
The first ball is dropped from rest, so its initial velocity (u1) = 0 m/s. We can use the equation of motion:
\[ h = u_1 t + \frac{1}{2} g t^2 \]
Substituting the values:
\[ 80 = 0 + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
Simplifying:
\[ 80 = 4.9 t^2 \]
\[ t^2 = \frac{80}{4.9} \approx 16.33 \]
\[ t \approx 4.03 \text{ seconds} \]

3. **Calculate the time taken by the second ball:**
The second ball is thrown downwards after 2 seconds, so:
\[ t_{second} = t - 2 \approx 4.03 - 2 = 2.03 \text{ seconds} \]

4. **Use the equation of motion for the second ball:**
Let the initial velocity of the second ball be \( u_2 \). The second ball is thrown downwards, so we can use the equation:
\[ h = u_2 t_{second} + \frac{1}{2} g t_{second}^2 \]
Substituting the values:
\[ 80 = u_2 \cdot 2.03 + \frac{1}{2} \cdot 9.8 \cdot (2.03)^2 \]
Calculate \( \frac{1}{2} \cdot 9.8 \cdot (2.03)^2 \):
\[ \frac{1}{2} \cdot 9.8 \cdot 4.1209 \approx 20.23 \]
Now substituting back:
\[ 80 = u_2 \cdot 2.03 + 20.23 \]
Rearranging gives:
\[ u_2 \cdot 2.03 = 80 - 20.23 \]
\[ u_2 \cdot 2.03 = 59.77 \]
\[ u_2 = \frac{59.77}{2.03} \approx 29.42 \text{ m/s} \]

5. **Final Result:**
The initial speed of the second ball is approximately **29.42 m/s**.