A cannon is placed on high ground 24 ft above a target that is 1000 ft away horizontally. It fires a projectile upward at a 45° angle with an initial speed of to ft/s. Whаt vаluе оf tip is necessary for the projectile to hit the target? The acceleration due to gravity is g = 32 ft/s2.

a. vo = (500)^1/2

b. vo = 150

с. vo = 100(2)^1/2

d. vo = 250

e. vo = 125(2)^1/2

yor answer is 7/6 times 5

step-by-step explanation:

[tex]Which of the following products is equivalent to the quotient 1 1/6 • 1/5[/tex]

D. vo = 125(2)^1/2

Step-by-step explanation:

Range of a projectile is the defined as the distance covered in the horizontal direction.

Range = U²sin2θ/g

U is the speed of the object

θ is the angle of launch

g is the acceleration due to gravity

If the target is 1000 ft away horizontally, range = 1000ft

θ = 45°

g = 32ft/s²

1000 = U²sin2(45°)/32

32000 =U²sin90°

32000 = U²

U = √32000

U = 125√2 m/s

(answer 2 )i’m very smart that’s how i know the answer.