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

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