The range of a projectile is modeled by the function , where v is the initial velocity and $\theta$ is the angle at which the object is initially propelled. The maximum range is achieved when $\theta$ = 45°. Use exact values to compute how many feet short of maximum the projectile falls if $\theta$ = 67.5° and v = 104 ft/sec.
A) feet
B) feet
C) feet
D) feet

Question

The range of a projectile is modeled by the function  , where v is the initial velocity and $\theta$ is the angle at which the object is initially propelled. The maximum range is achieved when $\theta$  = 45°. Use exact values to compute how many feet short of maximum the projectile falls if $\theta$  = 67.5° and v = 104 ft/sec.
A) feet 
B) feet 
C) feet 
D) feet

Quizplus · Accepted Answer

Using the provided function, we can calculate the maximum range by setting the angle to 45 degrees:

R(45) = v^2/g

where g is the acceleration due to gravity, approximately 32.2 ft/s^2. Plugging in v = 104 ft/s, we get:

R(45) = (104 ft/s)^2 / (32.2 ft/s^2) = 338.05 ft

To find how many feet short of the maximum range the projectile falls when launched at 67.5 degrees, we need to calculate the actual range R(67.5) and subtract the maximum range R(45):

R(67.5) = v^2 * sin(2θ) / g

Plugging in v = 104 ft/s and θ = 67.5 degrees, we get:

R(67.5) = (104 ft/s)^2 * sin(2*67.5°) / (32.2 ft/s^2) = 274.40 ft

The difference between the actual range and the maximum range is:

338.05 ft - 274.40 ft = 63.65 ft

So the projectile falls 63.65 feet short of the maximum range. The closest answer choice is (C) 11ea7f29_4ffd_2c5e_9ecd_4d9dd9011c73_TB3307_11 feet.

The Range of a Projectile Is Modeled by the Function θ\thetaθ

The Range of a Projectile Is Modeled by the Function $\theta$