I have a physics question

Question

A soccer ball is kicked vertically from the ground level with a speed of 20m/s. At what height is the gravitational potential energy of the ball maximum?

Answer 1

You can use the energy conservation law to do it.

Ep + Ek = constant

0 + (1/2)m v^2 = mgh + 0, for maximum potential energy

Solve for h,

h = v^2/(2g) = 20^2/(2*9.8) = 20.4 m

Answer: At h = 20.4 m the gravitaional potential energy of the ball reaches maximum.

Answer 2

P.E. is energy due to position

An easy kinematic equation can be used, and by taking it's 1st derivative and setting it = to 0, we can find the max height the ball is subjected to, and we can calculate that P.E.

1st eq; x= x₀ +v₀t + 1/2 at², a simple kinematic eq for distance. Next- take the1st derivative in terms of t, and set =0. We basically have a one-dimensional parabola we're finding the height of.

2nd eq (1st deriv.) - with some clean up -> v = v₀ + at. Next we are given an initial velocity, so solve for t and remember to set this 1st derivative =0  --->  0=20 m/s + (-9.8m *t) --clean up & t=2.04 seconds

Going back to eq #1 and remembering that at the top of the parabola the ball is still --> v=0

x= 0 + 0 + 1/2 (9.8m/s²) (2.04 s)², we get x = 20.4 m.

*I chose +g in the last eq.  because we were at a stop and there were no counter-forces (plus it gives a positive answer)

Answer 3

The highest the ball can go is the value that it has the largest gravitational potential energy. You may either calculate it outright , Vf2-Vo2 = 2gh or use the change in KE = the change in PE (1/2 m(Vf2-Vo2) = mgh ).

Either way the answer is the same, 20 m ( if you say g = 10 m/s2 0 or 20.4 m ( if g = 9.8 m/s2 ).

Answer 4

Since there are no retarding forces 

Mechanical energy is conserved.

 

KE1+PE1=KE2+PE2

 KE2=0 PE1=0 so we get KE1=PE2  1/2mv1^2=mgh2 so we get h2=0.5 v1^2/g=20.4 metersabove ground where potential energy is max.

Answer 5

H = V^2/2g

plug in values

H= 20*20/2*9.8 = 200/9.8 = 20.4 m

Answer 6

From the Law of Conservation of Energy, The total Mechanical energy will remain constant when air resistance , friction, etc. are not working,

So, 1/2 mv² (Initial Mechanical Energy) =mgH (Final Mechanical Energy)

or, H= v²/2g = 400/19.6 m = 20.4 m

Answer 7

Answer 8

The maximum potential energy of the ball will be reached at the point where the ball reaches its maximum height and stops (where v=0).

First find the time it takes to reach that height.

v_{f =} v_i + at

0 = 20 m/s + (-9.8 m/s²) (t)

-20 m/s = -9.8 m/s² (t)

t = 2.04 s

Knowing the time you can now find the maximum distance using

d = v_it + 1/2 at²

d = 20 m/s (2.04 s) + (-9.8 m/s²) (2.04s)²

d = 20.4m

Potential Energy = mad (mass x acceleration x distance) or mgh

P.E. = Mass (kg) X 408 m²/s² - Note the units will result in Newton-meters or Joules

Answer 9

The gravitational Potential Energy is maximum when the ball reaches its maximum height before turning to fall back.

A. One approach is to use conservation of energy principle.

When kicked up it has pure kinetic energy, E = mv²

When at the peak height, H, the ball speed is zero and the entire energy is potential energy, E=mgH

From conservation of energy, these two must be equal

1/2*mv² = mgH

Solving for H.

H = v²/2g

Here, v= 20m/s, g =9.8m/s²

Note the units. The result will be in meters.