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

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point

the correct answer is $7.38

step-by-step explanation: