7) a) horizontal component of the initial velocity is given by
V = UCos--------(1)
here from eqn (1) we have
U = V/Cos
= 18/Cos(65)
= 18/0.4226----(2)
U = 42.59m/s
vertical component of the initial velocity is
U' = V / Sin
=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
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