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3 years ago

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A ball of mass 0.152 kg is dropped from a height 2.27 m above the ground. The acceleration of gravity is 9.8 m/s 2 . Neglecting

air resistance, determine the speed of the ball when it is at a height 0.903 m above the ground. Answer in units of m/s.

2 answers:

VMariaS [17]3 years ago

7 0

Answer:

Height to achieve a speed of 2 m/s = 0.20m

Height to achieve a speed of 3 m/s = 0.46m

Height to achieve a speed of 4 m/s = 0.82m

Height to achieve a speed of 5 m/s = 1.28m

Height to achieve a speed of 6 m/s = 1.84m

Natali5045456 [20]3 years ago

5 0

Answer:

8.44 m/s.

Explanation:

Change in Potential Energy = Mass x Acceleration From Gravity x H2 - H1

Kinetic Energy = 1/2 (mass) x [(v2)^2 - v1^2]

g * h = 1/2 * v^2

(9.8) x (2.27) = 1/2 * (v)^2

v^2 = 2[(9.8) x (2.27)]

v = 6.67 m/s

g * delta h = 1/2 * delta v^2

(9.8) x (2.27 - 0.903) = 1/2 * [(v2)^2 - (6.67)^2]

v2^2 = 71.2821

v2 = 8.44 m/s

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To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

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Considering that the net strain energy of the sample is

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