The velocity with which the jumper leaves the floor is 5.1 m/s.
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What is the initial velocity of the jumper?</h3>
The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.
Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- v is the initial velocity of the jumper on the floor
- h is the maximum height reached by the jumper
- g is acceleration due to gravity
v = √(2 x 9.8 x 1.3)
v = 5.1 m/s
Learn more about initial velocity here: brainly.com/question/19365526
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To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.
First, we manipulate the one of the kinematic equations
v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped
Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,
<span>we get v0 = sqrt (2(a)(x))
Substituting the known values,
v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s
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Therefore, before stopping the car the original speed of the car would be 14.49 m/s
When the object is at the top of the hill it has the most potential energy. If it is sitting still, it has no kinetic energy. As the object begins to roll down the hill, it loses potential energy, but gains kinetic energy. The potential energy of the position of the object at the top of the hill is getting converted into kinetic energy. Hope this helped. :)
Answer:
Explanation:
Given
mass of sphere 
diameter of sphere 
radius 
friction will provide resisting torque so
where 
(b)time taken to decrease its rotational speed by 
