Complete Question:
A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level
a) Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.
b) What is the height, hn in meters?
Answer:
a) Energy = mghₙ
b) Height, hₙ = 5.02 m
Explanation:
a) Total energy in terms of maximum height
Let maximum height be hₙ
At maximum height, velocity, V=0
Total mechanical energy , E = mgh + 1/2 mV^2
Since V=0 at maximum height, the total energy in terms of maximum height becomes
Energy = mghₙ
b) Height, hₙ in meters
mghₙ = mgh + 1/2 mV^2
mghₙ = m(gh + 1/2 V^2)
Divide both sides by mg
hₙ = h + 0.5 (V^2)/g
h = 2.15m
g = 9.8 m/s^2
V = 7.5 m/s
hₙ = 2.15 + 0.5(7.5^2)/9.8
hₙ = 2.15 + 2.87
hₙ = 5.02 m
Answer:
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Answer:
<h2>
206.67N</h2>
Explanation:
The sum of force along both components x and y is expressed as;
The magnitude of the net force which is also known as the resultant will be expressed as 
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
Similarly,
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
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