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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?

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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
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.

David R. | Math, Physics, Chemistry, Music Tutor for HireMath, Physics, Chemistry, Music Tutor fo...
5.0 5.0 (1 lesson ratings) (1)
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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)

George M. | A very experienced math and science tutor.A very experienced math and science tuto...
4.9 4.9 (35 lesson ratings) (35)
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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 ).

Ali M. | Friendly and High Quality TutoringFriendly and High Quality Tutoring
4.8 4.8 (4 lesson ratings) (4)
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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.