A particle is projected with a velocity of 146 m/s at an elevation of 35°. Calculate:
the angle of projection at which the horizontal range and the maximum height of a projectile are
equal.

To find the angle of projection at which the horizontal range and the maximum height of a projectile are equal, we need to derive the necessary expressions and then equate them.

### Horizontal Range of a Projectile
The horizontal range \( R \) of a projectile is given by:

\[ R = \frac{v^2 \sin(2\theta)}{g} \]

where:
- \( v \) is the initial velocity of the projectile,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Maximum Height of a Projectile
The maximum height \( H \) of a projectile is given by:

\[ H = \frac{v^2 \sin^2(\theta)}{2g} \]

### Condition for Equal Range and Height
To find the angle where the horizontal range \( R \) is equal to the maximum height \( H \):

\[ R = H \]

Substituting the expressions for \( R \) and \( H \):

\[ \frac{v^2 \sin(2\theta)}{g} = \frac{v^2 \sin^2(\theta)}{2g} \]

Simplifying this:

\[ 2 \sin(2\theta) = \sin^2(\theta) \]

We know from the double-angle identity that:

\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]

So:

\[ 2 \times 2 \sin(\theta) \cos(\theta) = \sin^2(\theta) \]

\[ 4 \sin(\theta) \cos(\theta) = \sin^2(\theta) \]

Dividing both sides by \( \sin(\theta) \) (assuming \( \theta \neq 0 \)):

\[ 4 \cos(\theta) = \sin(\theta) \]

Dividing both sides by \( \cos(\theta) \):

\[ 4 = \tan(\theta) \]

Taking the arctangent of both sides:

\[ \theta = \tan^{-1}(4) \]

Thus, the angle of projection at which the horizontal range and the maximum height are equal is:

\[ \theta = \tan^{-1}(4) \approx 75.96^\circ \]

### Summary
The angle of projection at which the horizontal range and the maximum height of a projectile are equal is approximately \( 75.96^\circ \).

Would you like more details or have any questions?

Here are 5 questions you might want to ask next:
1. How do you derive the formulas for the horizontal range and maximum height of a projectile?
2. Can you explain the trigonometric identities used in solving this problem?
3. What happens to the projectile's range and height if the angle of projection is 45 degrees?
4. How does air resistance affect the horizontal range and maximum height of a projectile?
5. Can you provide examples of other projectile motion problems?

**Tip:** When dealing with projectile motion, breaking the problem into horizontal and vertical components can simplify the analysis and help you understand the motion better.

Calculate the angle of projection where the horizontal range and maximum height of a projectile are equal. Understand the formulas for projectile motion and trigonometric identities used in solving this problem. Suitable for advanced high school and college students.

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