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.