A stone thrown upward with a speed u from the top of a tower reaches the ground with a velocity 4u. The height of the tower is

Options

A. `(15u^(2))/(2g)`
B. `(7u^(2))/(2g)`
C. `(16u^(2))/(g)`
D. Zero

Answer

To solve the problem step by step, we will use the principles of kinematics.

### Step 1: Understand the scenario
A stone is thrown upwards with an initial velocity \( u \) from the top of a tower of height \( h \). After reaching its maximum height, it falls down and reaches the ground with a final velocity of \( 4u \).

### Step 2: Apply the kinematic equation
We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity = \( 4u \)
- \( u \) = initial velocity = \( u \)
- \( a \) = acceleration = \( -g \) (since gravity acts downwards)
- \( s \) = distance traveled = height of the tower \( h \)

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:
\[ (4u)^2 = u^2 + 2(-g)h \]
This simplifies to:
\[ 16u^2 = u^2 - 2gh \]

### Step 4: Rearrange the equation
Rearranging the equation to isolate \( h \):
\[ 16u^2 - u^2 = -2gh \]
\[ 15u^2 = -2gh \]
Now, we can solve for \( h \):
\[ h = \frac{15u^2}{2g} \]

### Step 5: Conclusion
Thus, the height of the tower \( h \) is given by:
\[ h = \frac{15u^2}{2g} \]