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
3.278*10^6 I think. Sorry if it’s wrong.
Between the top of the first and the top of the second loop, the coaster has lost potential energy = mgh, where h = 22.2 - 15 = 7.2m
This energy would have converted to Kinetic. Write out an equation and the masses will cancel out. Does that hint help you to find the solution? If not, I will give you another hint.
Answer:
C.) A high velocity and Large mass.
Explanation:
Momentum of any object is defined by following formula
Here
: m = mass of object
v = velocity of object
now we know that since momentum is product of mass and velocity
So in order to have more momentum we need the value of this product to be more. So this product will me large is both the physical quantity will be more in magnitude. So if mass is large and velocity will be more then the product of them will be large and hence the momentum of object will be more. Btw I had that question too.
