Physics Problem: Calculating Tower Height from Ball's Vertical Motion

This problem involves understanding the vertical motion of an object under gravity. We are given details about a ball thrown upwards from a tower, specifically its speed as it passes the starting point moving downwards and its speed upon hitting the ground. Our goal is to determine the height of the tower using these details and the acceleration due to gravity.

Understanding the Ball's Motion

Key information from the question:

• A ball is thrown vertically upwards from the top of a tower.
• It passes the point of projection (the top of the tower) while moving downwards with a speed of $20 \text{ m/s}$.
• It eventually hits the ground with a speed of $80 \text{ m/s}$.
• The acceleration due to gravity is given as $g = 10 \text{ m/s}^2$.

We need to find the height of the tower.

Method 1: Using Kinematic Equations

We can solve this problem by applying the equations of motion for constant acceleration. Let's establish a coordinate system:

• Let the upward direction be positive (+) and the downward direction be negative (-).
• The acceleration due to gravity acts downwards, so $a = -g = -10 \text{ m/s}^2$.
• We will consider the motion segment starting from when the ball passes the top of the tower moving downwards until it hits the ground.

Defining Variables for the Kinematic Calculation

• Initial Velocity ($u$): When the ball passes the tower top moving downwards, its velocity is $u = -20 \text{ m/s}$.
• Final Velocity ($v$): When the ball hits the ground, its velocity is $v = -80 \text{ m/s}$.
• Acceleration ($a$): As defined, $a = -10 \text{ m/s}^2$.
• Displacement ($s$): The displacement is from the top of the tower to the ground. Since the motion is downwards, the displacement is negative. Let the height of the tower be $h$. So, $s = -h$.

Applying the Relevant Kinematic Formula

The time-independent kinematic equation is suitable here:

$ v^2 = u^2 + 2as $

Calculating the Height of the Tower

Substitute the known values into the equation:

$ (-80)^2 = (-20)^2 + 2(-10)(-h) $

Calculate the squared terms:

$ 6400 = 400 + 20h $

Now, rearrange the equation to solve for $h$. First, subtract 400 from both sides:

$ 6400 - 400 = 20h $

$ 6000 = 20h $

Finally, divide by 20 to find the height:

$ h = \frac{6000}{20} $

$ h = 300 \text{ m} $

So, the height of the tower is 300 meters.

Method 2: Using Conservation of Energy

An alternative approach is to use the principle of conservation of mechanical energy. By symmetry, the speed with which the ball was initially thrown upwards from the tower top must be equal to the speed with which it passes the tower top moving downwards, which is $20 \text{ m/s}$.

Let's consider the energy changes from the moment the ball passes the tower top (moving downwards at $20 \text{ m/s}$) to the moment it hits the ground (moving at $80 \text{ m/s}$).

Energy Considerations

• Set the potential energy ($PE$) at the top of the tower to be zero.
• Initial State (at the tower top, moving down):
  • Velocity $v_i = 20 \text{ m/s}$
  • Kinetic Energy $KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (20)^2$
  • Potential Energy $PE_i = 0$
• Final State (at the ground):
  • Velocity $v_f = 80 \text{ m/s}$
  • Kinetic Energy $KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (80)^2$
  • Potential Energy $PE_f = -mgh$ (negative since the ground is below the reference point)

Applying the Conservation of Energy Principle

Total Initial Energy = Total Final Energy

$ KE_i + PE_i = KE_f + PE_f $

$ \frac{1}{2} m (20)^2 + 0 = \frac{1}{2} m (80)^2 - mgh $

Calculating the Height

Cancel the mass ($m$) from each term:

$ \frac{1}{2} (20)^2 = \frac{1}{2} (80)^2 - gh $

$ \frac{1}{2} (400) = \frac{1}{2} (6400) - gh $

$ 200 = 3200 - gh $

Rearrange to solve for $gh$:

$ gh = 3200 - 200 $

$ gh = 3000 $

Substitute the value of $g = 10 \text{ m/s}^2$:

$ (10)h = 3000 $

Solve for $h$:

$ h = \frac{3000}{10} $

$ h = 300 \text{ m} $

Final Answer: Height of the Tower

Both methods confirm that the height of the tower is 300 m.