a man hits a golf ball (0.2kg) which accelerates at a rate of 20 m/s what amount of force acted on the ball
2 answers:
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Answer:
Explanation:
<u>Horizontal Launch</u>
When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
vx=v
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
The horizontal component of the velocity is always the same:
The vertical component at t=5.5 s is:
Answer:
Explanation:
a ) Time period T = 2 s
Angular velocity ω = 2π / T
= 2π / 2 = 3.14 rad /s
Initial moment of inertia I₁ = 200 + mr²
= 200 + 25 x 2.5²
=356.25
Final moment of inertia
I₂ = 200 + 25 X 1.5 X 1.5
= 256.25
b ) We apply law of conservation of momentum
I₁ X ω₁ = I₂ X ω₂
ω₂ = I₁ X ω₁ / I₂
Putting the values
ω₂ = 4.365 rad s⁻¹
c ) Increase in rotational kinetic energy
=1/2 I₂ X ω₂² - 1/2 I₁ X ω₁²
.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²
= 684.95 J
This energy comes from work done against the centripetal pseudo -force.