You are throwing a stone straight-up in the absence of air friction. The stone is caught at the same height from which it was th
1 answer:
Answer:
A. True
Explanation:
When a stone is thrown straight-up, it has an initial velocity which decreases gradually as the stone move to maximum height due to constant acceleration due to gravity acting downward on the stone, at the maximum height the final velocity of the stone is zero. As the stone descends the velocity starts to increase and becomes maximum before it hits the ground.
Height of the motion is given by;
g is acceleration due to gravity which is constant
H is height traveled
u is the speed of throw, which determines the value of height traveled.
Therefore, when the stone is caught at the same height from which it was thrown in the absence of air resistance, the speed of the stone when thrown will be equal to the speed when caught.
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Answer:
1.23 m
Explanation:
The vertical distance covered by a free-falling object starting from rest in a time t is
where
g = 9.8 m/s^2 is the acceleration due to gravity
In this problem, we have
t = 0.50 s
So the distance covered is
To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.
The net height from the point where it begins to roll with an inclination of 30 degrees would be
In the case of Inertia would be given by
In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is
Replacing in Energy conservation Equation we have that
Potential Energy = Kinetic Energy of Rolling Object
Therefore the correct answer is C.