Question

A projectile is launched at an angle of 15 degrees above the horizontal and lands down range. For the same speed, what projection angle would produce the greatest downrange distance?

Answer 1

Answer:

45°

Explanation:

Range is the horizontal distance travelled by an object undergoing projectile motion.

Range is given by the formula:

$R=\frac{u^2sin(2\theta)}{g}$

where u = velocity, g = acceleration due to gravity and θ = angle above the horizontal.

For angle of 15°:

$R=\frac{u^2sin(2*15)}{g}=\frac{0.5u^2}{g}$

We get a maximum range when sin(2θ) = 1

sin(2θ) = 1

2θ = sin⁻¹(1)

2θ = 90⁰

θ = 90°/2

θ = 45⁰

For angle of 45°:

$R=\frac{u^2sin(2*45)}{g}=\frac{u^2}{g}$