Understanding How Kinetic Energy Reduces in Projectile Motion

When a body is thrown upwards with a certain initial velocity, its kinetic energy starts decreasing as it gains height. This is because the initial kinetic energy is being converted into potential energy due to gravity. The question asks for the specific height where the kinetic energy reduces to half of its initial value.

We are given the initial velocity (\(v_0\)) as 4 m/s and the acceleration due to gravity (\(g\)) as 10 m/s².

Calculating the Initial Kinetic Energy (KE₀)

The initial kinetic energy (\(KE_0\)) of the body when it is thrown upwards with an initial velocity \(v_0\) is given by the formula:

\(KE_0 = \frac{1}{2} m v_0^2\)

Where \(m\) is the mass of the body. Substituting the given initial velocity \(v_0 = 4\) m/s:

\(KE_0 = \frac{1}{2} m (4 \, \text{m/s})^2 = \frac{1}{2} m (16 \, \text{m}^2/\text{s}^2) = 8m \, \text{J}\)

This is the initial kinetic energy of the body.

Finding the Height Where Kinetic Energy Reduces to Half

We want to find the height (\(h\)) where the kinetic energy (\(KE_h\)) is half of the initial kinetic energy (\(KE_0\)).

\(KE_h = \frac{1}{2} KE_0\)

Substituting the value of \(KE_0\):

\(KE_h = \frac{1}{2} (8m \, \text{J}) = 4m \, \text{J}\)

At this height \(h\), let the velocity of the body be \(v_h\). The kinetic energy at height \(h\) is also given by:

\(KE_h = \frac{1}{2} m v_h^2\)

So, we have:

\(\frac{1}{2} m v_h^2 = 4m\)

We can cancel the mass \(m\) from both sides:

\(\frac{1}{2} v_h^2 = 4\)

\(v_h^2 = 8\)

So, the velocity of the body at the height where its kinetic energy reduces to half is \(v_h = \sqrt{8}\) m/s.

Applying Energy Conservation in Projectile Motion

We can use the principle of energy conservation to find the height \(h\). According to the law of energy conservation, the total mechanical energy (sum of kinetic and potential energy) remains constant if only conservative forces (like gravity) are doing work. In this case of projectile motion, mechanical energy is conserved.

Initial total energy (at height \(h=0\)) = Final total energy (at height \(h\))

\(KE_0 + PE_0 = KE_h + PE_h\)

Where \(PE_0\) is the initial potential energy and \(PE_h\) is the potential energy at height \(h\). We take the initial height as zero, so \(PE_0 = mg(0) = 0\). The potential energy at height \(h\) is \(PE_h = mgh\). The kinetic energy reduces to half, so \(KE_h = \frac{1}{2} KE_0\).

Detailed Calculation Using Energy Conservation

Using the energy conservation equation:

\(KE_0 + 0 = \frac{1}{2} KE_0 + mgh\)

Subtract \(\frac{1}{2} KE_0\) from both sides:

\(KE_0 - \frac{1}{2} KE_0 = mgh\)

\(\frac{1}{2} KE_0 = mgh\)

We already calculated \(KE_0 = 8m\). Substituting this value:

\(\frac{1}{2} (8m) = mgh\)

\(4m = mgh\)

Cancel the mass \(m\) from both sides:

\(4 = gh\)

Now, substitute the value of \(g = 10\) m/s²:

\(4 = (10 \, \text{m/s}^2) h\)

Solving for \(h\):

\(h = \frac{4}{10} \, \text{m}\)

\(h = 0.4 \, \text{m}\)

Summary for Projectile Motion

The height at which the kinetic energy reduces to half of its initial value is 0.4 m. This demonstrates the conversion between initial kinetic energy and potential energy as the body moves against gravity in projectile motion, following the principle of energy conservation. The height where the kinetic energy reduces depends on the initial velocity and the acceleration due to gravity.