Question

A football is kicked with a speed of 18 m/sec an angle of 65 degrees to horizontal. What are the horizontal and vertical components of the initial velocity of football. How long is football in air How far does football travel horizontally before it hits the ground. A baseball player bounces off a baseball off the floor by giving it a downward velocity of 3 m/sec at height of 1 m above the floor. The ball bounces off the floor and rises to the height of 0.8 m. What is the impulse applied by the floor to the ball, find the magnitude and direction Mass of ball is 0.5 kg.

Answer 1

7) a) horizontal component of the initial velocity is given by

V = UCos$\Theta$--------(1)

here from eqn (1) we have

U = V/Cos$\Theta$

   = 18/Cos(65)

   = 18/0.4226----(2)

U = 42.59m/s

vertical component of the initial velocity is

U' = V / Sin$\Theta$

   =18/Sin(65)

=18/0.9060

=19.86m/s----------(3)

b) average velocity along the horizontal and vertical is given by

U1 = (U + U') / 2-------(4)

      = (42.59 + 19.86)/2

U1 = 31.22m/s------(5)

V = U1 -gt-------(6)

18 = 31.22 - (9.8)(t)

t = 13.22 / 9.8 = 1.34sec

football takes 1.34sec in air

c) V^2 = U^2 + 2gS

      (18)^2     = (31.22)^2 + 2(9.8)S

S = 33.19cm

8) the ball falls vertically through a distance

S = 1-0.8 = 0.2m

let 'V' be the velocity at the time of falling WHEN U =0

V^2 = U^2 + 2gS

V^2 = 0 + 2(9.8)(0.2)

V = 1.97m/s

velocity V = S/t

3 = 0.2/t

t = 0.06sec

acceleration a = (V - U)/t

a = (3-1.97)/0.06

= 17.16m/s^2

force   F = mxa

             = (0.5)(17.16)

    F = 8.58N

IMPULSE= Ft

           = (8.58)(0.06)

        = 0.514Ns