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
Gravitational Potential Energy GPE = mg∆h. 12. A 5.0 kg mass is initially sitting on the floor when it is lifted onto a table 1.15 meters high at.
Answer:
F = -307.4 N
Explanation:
It is given that,
Mass of the baseball, m = 0.145 kg
Initial speed of the baseball, u = 60 m/s
Final speed of the baseball, 
Time of contact, 
(a) It is assumed to find the horizontal component of average force. It is given by :
F = -307.4 N
So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.
Answer:
Two forces that act in opposite directions produce a resultant force that is smaller than either individual force. To find the resultant force subtract the magnitude of the smaller force from the magnitude of the larger force. The direction of the resultant force is in the same direction as the larger force.
The answer to that would be that
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