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A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the
horizontal. Determine the time of flight, the horizontal displacement, and the peak
height of the football. (Remember the velocity at peak height)

Answer :

Answer:

a

Explanation:

temdan2001

With the use of formula, the time of flight, the horizontal displacement, and the peak height of the football are 1.8s, 63.6 m, 15.8 m

PROJECTILE

The object that is been projected to take a trajectory path is known as Projectile.

Given that a football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal.

a. The time of flight can be calculated by using the formula below

t = usinФ / g

Where

u = initial velocity = 25 m/s

Ф = 45 degrees

Substitute all the parameters into the formula

t = (25 x sin 45) / 9.8

t = 17.678 / 9.8

t = 1.8 s

b. The horizontal displacement R, will be

R = ucosФT

Where T = 2t = 2 x 1.8 = 3.6s

R = 25cos45 x 3.6

R = 63.64 m

c. The peak height of the football.

At maximum height, V = 0

H = [tex]U^{2}[/tex][tex]Sin^{2}[/tex]Ф / 2g

H = [tex]25^{2}[/tex] [tex](Sin45)^{2}[/tex] / (2 x 9.8)

H = 312.5 / 19.8

H = 15.78 m

Therefore, the time of flight, the horizontal displacement, and the peak height of the football are 1.8s, 63.6 m, 15.8 m approximately.

Learn more about Projectile here: https://brainly.com/question/24216590

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