The potential energy of a projectile at its maximum height is equal to its kinetic energy there. Its range for velocity of projection u is

Answer

To solve the problem, we need to find the range of a projectile when its potential energy at maximum height is equal to its kinetic energy at that height. Let’s break down the solution step by step.

### Step 1: Understand the situation
When a projectile is launched with an initial velocity \( u \) at an angle \( \theta \), it has both vertical and horizontal components of velocity. The vertical component is \( u \sin \theta \) and the horizontal component is \( u \cos \theta \).

### Step 2: Determine the maximum height
The maximum height \( h \) of the projectile can be calculated using the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Calculate the kinetic energy at maximum height
At maximum height, the vertical component of velocity is zero, and the only velocity is the horizontal component \( u \cos \theta \). The kinetic energy \( KE \) at this point is given by:
\[ KE = \frac{1}{2} m (u \cos \theta)^2 = \frac{1}{2} m u^2 \cos^2 \theta \]

### Step 4: Calculate the potential energy at maximum height
The potential energy \( PE \) at maximum height is given by:
\[ PE = mgh = mg \left(\frac{u^2 \sin^2 \theta}{2g}\right) = \frac{1}{2} m u^2 \sin^2 \theta \]

### Step 5: Set potential energy equal to kinetic energy
According to the problem, the potential energy at maximum height is equal to the kinetic energy at that height:
\[ PE = KE \]
Substituting the expressions we derived:
\[ \frac{1}{2} m u^2 \sin^2 \theta = \frac{1}{2} m u^2 \cos^2 \theta \]

### Step 6: Simplify the equation
We can cancel \( \frac{1}{2} m u^2 \) from both sides (assuming \( m \) and \( u \) are not zero):
\[ \sin^2 \theta = \cos^2 \theta \]

### Step 7: Use the identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can rewrite:
\[ \sin^2 \theta = \cos^2 \theta \implies \tan^2 \theta = 1 \implies \tan \theta = 1 \]
Thus, \( \theta = 45^\circ \).

### Step 8: Calculate the range
The range \( R \) of the projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting \( \theta = 45^\circ \):
\[ R = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g} \]

### Final Answer
Thus, the range of the projectile when its potential energy at maximum height equals its kinetic energy at that height is:
\[ R = \frac{u^2}{g} \]