Answer:
10.77kgm/s
Explanation:
Impulse is defined as change in momentum of a body.
From Newton's second law
F = ma
F = m(v-u)/t
Ft = m(v-u) = Impulse
where;
M is the mass of the object
v is the final velocity of the object
u is the initial velocity of the object.
Using the formula to time the impulse
Impulse = m(v-u)
m = 0.745kg
Since the ball dropped from rest at a height of 3.75m, it's initial velocity can be calculated according to the equation of motion
g is positive (since it dropped from a height I.e downward motion)
v²= u²+2gH
v² = 0²+2(9.8)(3.75)
v² = 73.5
v = √73.5
v = 8.57m/s(initial velocity of the ball)
If the ball rebounds to a height if 1.78m, g will be negative (since it is an upward motion)
In this case
v² = u²-2gH
0² = u²-2(9.8)(1.78)
-u² = -34.89
u = √34.89
u = 5.91m/s (final velocity of the ball)
Impulse = 0.745(5.91-(-8.57))
Impulse = 0.745(5.91+8.57)
Impulse = 0.745(14.48)
Impulse = 10.77kgm/s
Note that the initial velocity was negated while calculating impulse due to rebounce.
