At time t = 0, a football player kicks a ball from the point a with position vector (2i + j) m on a horizontal football field. the motion of the ball is modelled as that of a particle moving horizontally with constant velocity (5i + 8j) m s–1.

find

(a) the speed of the ball,

(b) the position vector of the ball after t seconds.

the point b on the field has position vector (10i + 7j) m.

(c) find the time when the ball is due north of b.

at time t = 0, another player starts running due north from b and moves with constant speed v m s–1. given that he intercepts the ball,

(d) find the value of v.

(e) state one physical factor, other than air resistance, which would be needed in a refinement of the model of the ball’s motion to make the model more realistic.

a) 9.434 m/s

b) i (2+5*t) + (1+8*t-4.905*t²) j

c) t= 8/5 secs

d) 3.598 m/s

e) See explanation

Step-by-step explanation:

Part a)

The speed of the ball can be calculated from the given velocity v = 5i +8j

Taking magnitude of v = 5i + 8j

magnitude (v) = = 9.434 m/s.

Part b)

Using kinematic equation of particle as follows:

Sf = Si + Vi*t + 0.5*a*t² ..... Eq 1

Given: Si = (2i + j) m ; Vi = (5i+8j) m/s; a = -9.81 j m/s²

We evaluate Eq 1:

Sf = (2i+j) + (5i+8j)*t + 0.5*(-9.81j)*t²

We get after combining similar terms:

Sf = i (2+5*t) + (1+8*t-4.905*t²) j ..... Eq 2

Part c)

Using kinematic equation of particle only in i axis as follows we use Eq 1:

Sf = Si + Vi*t + 0.5*a*t²

Given: Si = 2 m ; Sf = 10; Vi = 5 m/s; a = 0;

We evaluate Eq 1:

10 = 2 + 5*t - Solve for t

t = 8/5 seconds

Note: The above is the time t when the ball is due north of (10i+7j) i.e having a position vector of 10 in east direction but unknown in north direction. A point directly above or below 10i + 7j.

Part d)

The interception of ball and the player occurs at the same t = 8/5 secs and @ position vector (10i + aj) where a is a constant needs to be found.

Find a:

Using Eq 2 found in part b:

Sf = i (2+5*t) + (1+8*t-4.905*t²) j

Evaluate @ t= 8/5 secs

Sf = (10) i + (1.2432) j .... Eq 3

To find the speed v of the player when he intercepts the ball at Sf = (10) i + (1.2432) j is evaluated as follows:

v = change in position of player / Time

Hence, v = -3.598 j = 3.598 m/s

Part e)

Friction between the ball and surface from which is launched.

reduce the friction between its moving parts

step-by-step explanation:

in fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00