A Foot ball is at rest, a player kicks them for 0.3 sec andso ball attains velocity of 180km/hr. If mass of ball is 10g calculate force applied by footballer in SI system
Answers
Answer:
Motion, in physics, change with time of the position or orientation of a body. ... Motion that changes the orientation of a body is called rotation. In both cases all points in the body have the same velocity (directed speed) and the same acceleration (time rate of change of velocity).
(a) We take the force to be in the positive direction, at least for earlier times. Then the impulse is
J=∫
0
3.0×10
−3
Fdt=∫
0
3.0×10
−3
[(6.0×10
6
)t−(2.0×10
9
)t
2
]dt
=[
2
1
(6.0×10
6
)t
2
−
3
1
(2.0×10
9
)t
3
]
∣
∣
∣
∣
∣
∣
∣
∣
0
3.0×10
−3
=9.0N⋅s
(b) Since J=F
avg
Δt , we find
F
avg
Δt
J
=
3.0×10
−3
s
9.0N⋅s
=3.0×10
3
N .
(c) To find the time at which the maximum force occurs, we set the derivative of F with respect to time equal to zero, and solve for t. The result is t=1.5×10
−3
s . At that time the force is
F
max
=(6.0×10
6
)(1.5×10
−3
)−(2.0×10
9
)(1.5×10
−3
)
2
=4.5×10
3
N .
(d) Since it starts from rest, the ball acquires momentum equal to the impulse from the kick. Let m be the mass of the ball and ν its speed as it leaves the foot. Then,
ν=
m
o
=
m
J
=
0.45kg
9.0N⋅s
=20m/s
The force as function of time is shown below. The area under the curve is the impulse J. From the plot, we readily see that F(t) is a maximum at t=0.0015s , with F
max
=4500N
