A person, standing on the roof of a 40 m high tower, throws a ball vertically upwards with speed 10 m/s. Two seconds later, he throws another ball again in vertical direction (use `g=10(m)/(s^2)`). Both the balls hit the ground simultaneously.

Options

A. The first stone hits the ground after `4` seconds.
B. The second ball was projected vertically downwards with speed `5 m//s`.
C. The distance travelled by the first ball is `10 m` greater than the distance travelled by the second ball.
D. Both ball hit the ground with same velocities.

Answer

To solve the problem, we need to analyze the motion of both balls thrown from the tower. We will use the equations of motion under uniform acceleration due to gravity.

### Step 1: Analyze the first ball's motion
The first ball is thrown upwards with an initial velocity \( u_1 = 10 \, \text{m/s} \) from a height of \( h = 40 \, \text{m} \). We need to find the time it takes for this ball to hit the ground.

Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s = -40 \, \text{m} \) (the displacement is negative because it falls down), \( u = 10 \, \text{m/s} \), and \( a = -g = -10 \, \text{m/s}^2 \).

Substituting the values:
\[ -40 = 10t - \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ -40 = 10t - 5t^2 \]
Rearranging gives:
\[ 5t^2 - 10t - 40 = 0 \]
Dividing through by 5:
\[ t^2 - 2t - 8 = 0 \]

### Step 2: Solve the quadratic equation
To solve the quadratic equation \( t^2 - 2t - 8 = 0 \), we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -2, c = -8 \):
\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]
\[ t = \frac{2 \pm \sqrt{4 + 32}}{2} \]
\[ t = \frac{2 \pm \sqrt{36}}{2} \]
\[ t = \frac{2 \pm 6}{2} \]
This gives us two possible solutions:
\[ t = \frac{8}{2} = 4 \, \text{s} \quad \text{and} \quad t = \frac{-4}{2} = -2 \, \text{s} \quad (\text{not valid}) \]
Thus, the first ball takes \( t_1 = 4 \, \text{s} \) to hit the ground.

### Step 3: Analyze the second ball's motion
The second ball is thrown 2 seconds after the first ball, so it is thrown at \( t = 2 \, \text{s} \) and will hit the ground at \( t = 4 \, \text{s} \), meaning it is in motion for \( t_2 = 4 - 2 = 2 \, \text{s} \).

Let the initial velocity of the second ball be \( u_2 \). Using the same equation of motion:
\[ s = u_2 t_2 + \frac{1}{2} (-g) t_2^2 \]
Again, \( s = -40 \, \text{m} \), \( t_2 = 2 \, \text{s} \), and \( g = 10 \, \text{m/s}^2 \):
\[ -40 = u_2 \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2^2) \]
\[ -40 = 2u_2 - 20 \]
Rearranging gives:
\[ 2u_2 = -40 + 20 \]
\[ 2u_2 = -20 \]
\[ u_2 = -10 \, \text{m/s} \]
This indicates the second ball is thrown downwards with a speed of \( 10 \, \text{m/s} \).

### Conclusion
Both balls hit the ground simultaneously, with the first ball taking 4 seconds and the second ball taking 2 seconds (after being thrown). The initial speed of the second ball is \( 10 \, \text{m/s} \) downwards.