Question

The range r and the maximum height h of a projectile fired at an inclination θ to the horizontal with initial speed v0 are given by r(θ)= v20sin(2θ) g and h(θ)= v20(sinθ)2 2g​, where g≈32.2 feet per second per second is the acceleration due to gravity. find the range and maximum height of a projectile fired at an angle of 45° to the horizontal with an initial speed of 70 feet per second. question content area bottom part 1 r=enter your response here ▼ feet feet per second per second feet per second ​(round to two decimal places as​ needed.)

What is the range of the projectile?

A) 100.00 feet

B) 70.00 feet

C) 50.00 feet

D) 30.00 feet

Answer 1

The range of the projectile fired at an angle of 45° with an initial speed of 70 feet per second is 100 feet.

Step-by-step explanation:

The range of a projectile fired at an angle of 45° to the horizontal with an initial speed of 70 feet per second can be found by substituting the given values into the range formula:

R = v0^2sin(2θ)/g

R = (70^2)sin(2(45))/32.2

R = 100 feet

Therefore, the range of the projectile is 100 feet.