Problema Solution
A projectile is fired with an initial velocity 500 feet per second at an angle of 45 degrees with the horizontal. To the nearest 10 feet, find the horizontal distance covered by the projectile.
Answer provided by our tutors
let d = the horizontal distance that is the range of the projectile that we need to dins
d = (1/g) (v^2) * sin(2 * angle A)
g = 32 f/s^2: the gravitational acceleration
angle A = 45: the angle at which the projectile is launched
v = 500 fps: the velocity at which the projectile is launched
d = (1/32)*(500^2)*sin(2*45)
d = 7,812.5 ft or 7810 ft to the nearest 10 feet
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