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4 Answers

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.

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)

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 ).

Since there are no retarding forces 

Mechanical energy is conserved.



 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.