From the top of a tower, a stone A is thrown upwards and a stone B is thrown downwards with the same speed. The velocity of stone A, on colliding with the ground is :

Options

A. Greater than the velocity of B
B. Less than the velocity of B
C. The velocities of stones A and B will be same
D. Both the stones will fall on the earth at the same time

Answer

To find the velocity of stone A when it collides with the ground after being thrown upwards, we can use the kinematic equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the variables
- Let \( u \) be the initial velocity of both stones (upwards for stone A and downwards for stone B).
- Let \( h \) be the height of the tower.
- The acceleration due to gravity \( g \) acts downwards.

### Step 2: Write the equation of motion for stone A
When stone A is thrown upwards, it will first move upwards and then come down. The displacement when it hits the ground is \( -h \) (downward).

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s = -h \), \( u = u \), and \( a = -g \) (since gravity acts downward).

So, we can write:
\[ -h = ut - \frac{1}{2} g t^2 \]

### Step 3: Use the equation for final velocity
We can also use the equation that relates initial velocity, final velocity, acceleration, and displacement:
\[ v^2 = u^2 + 2as \]
For stone A:
- \( s = -h \)
- \( a = -g \)

Substituting these values into the equation gives:
\[ v^2 = u^2 + 2(-g)(-h) \]
This simplifies to:
\[ v^2 = u^2 + 2gh \]

### Step 4: Solve for the final velocity \( v \)
Taking the square root of both sides, we find:
\[ v = \sqrt{u^2 + 2gh} \]

### Conclusion
The velocity of stone A when it collides with the ground is:
\[ v = \sqrt{u^2 + 2gh} \]

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