A ball is kicked from a location <8, 0, -7> (on the ground) with initial velocity <-8, 19, -3> m/s. The ball's speed is low enough that air resistance is negligible.(a) What is the velocity of the ball 0.3 seconds after being kicked? (Use the Momentum Principle!)(b) What is the average velocity of the ball over this time interval?(c) Use the average velocity to find the location of the ball 0.5 seconds after being kicked.

Respuesta :

Answer:

a) v(0.3) = ( 8, 0 , 9.943 )

b) 21.1 m/s

c) S(0.5) = (4, 9.5 , 9.72625)

Explanation:

s(0) = 8i + 0j -7k

v(0) = -8i +19j -3k

a(t) = (0i+0j-9.81k)

part a

v(t) = v(0) + a(t)*t

v(0.3) = 8i + 0j -7k + (0i+0j-9.81k)* 0.3

v(0.3) = 8i + 0j - 9.943k

part b

S(t) = s(0) + v(0)*t + 0.5*a(t)*t^2

S(0.3) - S (0) = v(0)*t + 0.5*a(t)*t^2

= (-8i +19j -3k)*0.3 + 0.5*(0i+0j-9.81k)*(0.3)^2

= (-2.4i + 5.7j - .9k) + (0i+0j-0.44145k)

Displacement = (-2.4i + 5.7j - 1.34145k)

[tex]distance = \sqrt{2.4^2+5.7^2+1.34145^2}\\ = 6.3285 m[/tex]

Average velocity = 6.3285 / 0.3 = 21.1 m/s

part c

S(t) = s(0) + v(0)*t + 0.5*a(t)*t^2

S(0.5) = v(0)*t + 0.5*a(t)*t^2 + S(0)

=(-8i +19j -3k)*0.5 + 0.5*(0i+0j-9.81k)*(0.5)^2 + (8i + 0j -7k)

=(-4i + 9.5j - 1.5k) + (0i+0j-1.22625k) + (8i + 0j -7k)

=(4i + 9.5j - 9.72625k)

In this exercise we have to use the knowledge of vectors to calculate the speed of the ball, so:

A) [tex]v(0.3) = ( 8, 0 , 9.943 )[/tex]

B) [tex]21.1 m/s[/tex]

C) [tex]S(0.5) = (4, 9.5 , 9.72625)[/tex]

Given the following information about the equations we have:

  • [tex]s(0) = 8i + 0j -7k[/tex]
  • [tex]v(0) = -8i +19j -3k[/tex]
  • [tex]a(t) = 0i+0j-9.81k[/tex]

A)So calculating the speed of the ball, we have:

[tex]v(t) = v(0) + a(t)*t\\v(0.3) = 8i + 0j -7k + (0i+0j-9.81k)* 0.3\\v(0.3) = 8i + 0j - 9.943k[/tex]

B) So calculating the average velocity we have:

[tex]S(t) = s(0) + v(0)*t + 0.5*a(t)*t^2\\S(0.3) - S (0) = v(0)*t + 0.5*a(t)*t^2\\= (-8i +19j -3k)*0.3 + 0.5*(0i+0j-9.81k)*(0.3)^2\\= (-2.4i + 5.7j - .9k) + (0i+0j-0.44145k)\\= 6.3285 / 0.3 = 21.1 m/s[/tex]

C) Then calculating the average velocity of the ball in 0.5 seconds, we find that:

[tex]S(t) = s(0) + v(0)*t + 0.5*a(t)*t^2\\S(0.5) = v(0)*t + 0.5*a(t)*t^2 + S(0)\\=(-8i +19j -3k)*0.5 + 0.5*(0i+0j-9.81k)*(0.5)^2 + (8i + 0j -7k)\\=(-4i + 9.5j - 1.5k) + (0i+0j-1.22625k) + (8i + 0j -7k)\\=(4i + 9.5j - 9.72625k)[/tex]

See more about vectores at brainly.com/question/13188123