Question

The range of a projectile that is launched with an initial velocity v at an angle of theta with the horizontal is given by R = v^2/g sin 2theta, where g is the acceleration due to gravity or 9.8 meters per second squared. If a projectile is launched with an initial velocity of 15 meters per second, what angle is required to achieve a range of 20 meters? Round answers to the nearest whole number.

a) 12 degrees
b) 30 degrees
c) 60 degrees
d) 75 degrees

Answer 1

To achieve a range of 20 meters with an initial velocity of 15 m/s, an angle of approximately 12 degrees is required.

Step-by-step explanation:

The range of a projectile launched at an angle above the horizontal with an initial velocity v can be calculated using the formula R = v^2/g sin 2θ, where g is the acceleration due to gravity. In this case, we are given an initial velocity of 15 m/s and a desired range of 20 m. To find the angle required to achieve this range, we can rearrange the formula as follows:

20 = (15^2/9.8) sin 2θ

Using the inverse sine function, we can find the value of 2θ:

sin^(-1)(20*9.8/15^2) = 2θ

Simplifying the equation gives:

θ ≈ 12°