1. Which angle results in the widest range?
The angle that results in the widest range for a projectile, assuming all other conditions remain constant (such as initial velocity, launch height, and no air resistance), is 45°. The reason for this is that the range R of a projectile is given by the formula:
\[ R = \frac{v^2 \cdot \sin(2\theta)}{g} \]
where \( v \) is the initial velocity, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity. The range R is maximized when \( \sin(2\theta) \) is maximized, which occurs when \( 2\theta = 90° \) or, equivalently, when \( \theta = 45° \), because \( \sin(90°) = 1 \) is the maximum value of the sine function.
2. Which angle results in a maximum height?
The maximum height \( H \) of a projectile is achieved when the launch angle \( \theta \) results in the highest vertical component of the initial velocity. The vertical component is given by \( v \cdot \sin(\theta) \), and since the sine function reaches its maximum at 90°, the angle closest to 90° will result in the highest vertical component and thus the maximum height. Given your options, 75° is the angle that results in the maximum height because it yields the greatest upward component compared to the other angles listed. The maximum height \( H \) is given by:
\[ H = \frac{(v \cdot \sin(\theta))^2}{2g} \]
By increasing \( \theta \) towards 90°, \( v \cdot \sin(\theta) \) and thus \( H \) are increased.
3. How would you compare the distance traveled by a projectile launched at an angle of 15° and 75°? Explain why it happens.
Projectiles launched at angles of 15° and 75° actually travel the same horizontal distance under ideal conditions (no air resistance and assuming other factors like launch speed and height are constant). This is because the projectile range equation involves the sine of twice the launch angle (as shown earlier), and \( \sin(2\theta) \) is symmetric about 45° for the angles within the first quadrant (0° to 90°).
Mathematically, since the sine function is periodic and \( \sin(\alpha) = \sin(180° - \alpha) \), we have:
\[ \sin(2 \cdot 15°) = \sin(30°) \]
\[ \sin(2 \cdot 75°) = \sin(150°) \]
Both \( \sin(30°) \) and \( \sin(150°) \) are equal. Hence, both angles when doubled are equivalent in terms of their contribution to the range formula, which means:
\[ R_{15°} = R_{75°} \]
This is because the range equation depends only on the sine value, which is the same for both angles once doubled due to the supplementary angle identity of sine. So, a projectile launched at 15° and another at 75° will have the same horizontal range when other factors are equal.
