Q.A cricket bowler releases the ball in two different ways

The key idea here is the Conservation of Mechanical Energy in projectile motion when air resistance is negligible.

1. After release, the only force acting on the ball is gravity. Since gravity is a conservative force and no non-conservative forces (like air resistance) are doing work, the total mechanical energy of the ball remains constant throughout its flight.
2. The initial mechanical energy $E_i$ at height $H$ is the sum of its initial kinetic energy and potential energy: $E_i = \frac{1}{2}mv_s^2 + mgH$. Here, $m$ is the mass of the ball, $v_s$ is its initial speed, $g$ is the acceleration due to gravity, and $H$ is the initial height.
3. When the ball hits the ground (height $0$), its final mechanical energy $E_f$ is purely kinetic: $E_f = \frac{1}{2}mv_f^2$, where $v_f$ is the final speed. …

When a ball is released and moves under the influence of gravity alone (neglecting air resistance), its total mechanical energy remains constant. This principle is key to understanding how its speed changes. Mechanical energy is the sum of kinetic energy (energy due to motion) and potential energy (energy due to position).

The crucial insight here is that the initial speed $v_s$ determines the initial kinetic energy, regardless of the direction of the initial velocity. Since the initial height $H$ is also the same for both scenarios, the initial potential energy is identical. As mechanical energy is conserved, the final kinetic energy (and thus the final speed) must also be the same when the ball reaches the ground.

Let's break this down step-by-step.

1. Identify the governing principle:

   The problem describes the motion of a ball under gravity, with air resistance neglected. In such a situation, the only force doing work is gravity, which is a conservative force. This means that the total mechanical energy of the ball remains constant throughout its flight.

   The principle of conservation of mechanical energy states:

   $KE_i + PE_i = KE_f + PE_f$

   where $KE = \frac{1}{2}mv^2$ is kinetic energy and $PE = mgh$ is gravitational potential energy.

2. Define initial and final conditions:

   Let $m$ be the mass of the ball.

   • Initial conditions:
     • Initial speed: $v_i = v_s$ (given as the same for both scenarios).
     • Initial height: $h_i = H$ (given as the same for both scenarios).
   • Final conditions:
     • Final height: $h_f = 0$ (when the ball hits the ground).
     • Final speed: $v_f$ (this is what we need to find and compare).

3. Apply the conservation of mechanical energy equation:

   Substitute the expressions for kinetic and potential energy, along with the initial and final conditions, into the conservation of mechanical energy formula:

   $\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$

   Substituting the specific values:

   $\frac{1}{2}mv_s^2 + mgH = \frac{1}{2}mv_f^2 + mg(0)$

   This simplifies to:

   $\frac{1}{2}mv_s^2 + mgH = \frac{1}{2}mv_f^2$

4. Solve for the final speed $v_f$:

   We can cancel out the mass $m$ from every term in the equation:

   $\frac{1}{2}v_s^2 + gH = \frac{1}{2}v_f^2$

   Now, multiply the entire equation by 2 to clear the fractions:

   $v_s^2 + 2gH = v_f^2$

   Finally, take the square root to find the final speed (speed is always a positive value):

   $v_f = \sqrt{v_s^2 + 2gH}$

5. Compare the final speeds for both scenarios: …

Concept: Conservation of Mechanical Energy

With air resistance neglected, only gravity (a conservative force) does work, so total mechanical energy is conserved — the final speed depends only on the initial speed and height, not on the launch direction.

Step 1: Write the initial mechanical energy

Step 2: Write the final mechanical energy (just before impact, height = 0)

Step 3: Apply conservation of energy

$\tfrac12 mv_s^2+mgH=\tfrac12 mv_f^2 \implies v_f=\sqrt{v_s^2+2gH}$ …