I'm stumped because there's no indication of whether the ball was thrown from the ground or what... Please help!
A ball of mass m is thrown straight up and reaches height h above the ground. Let the ground be where height is defined to be zero. There is no friction or air resistance. At the original point where the ball is thrown, which of the following give the correct expressions for the ball's mechanical energy (ME), kinetic energy (KE), and potential energy (PE)?
(A) ME=0
KE=0
PE=mgh
(B) ME=mgh
KE=.5mgh
PE=.5mgh
(C) ME=mgh
KE=mgh
PE=0
(D) ME=mgh
KE=0
PE=mgh
(E) ME=mgh +.5mv^2
KE=.5mv^2
PE=0
The test is tomorrow, this is a practice test which is usually similar to our tests. Any help is greatly appreciated... and I don't just want the answer because that's really useless. Thank you!!!
Answers & Comments
Verified answer
I think the first two sentences effectively tell you that the ball was thrown from the ground.
At the point where the ball is thrown, it has KE but no PE. Therefore, (C).
40 8. paintings is only Fx d the place d is interior the process the tension. The paintings performed is only mgh. h = a million/2 *60m = 30 m W = 70 kg * 9.80 one m/s^2 * 30m = 20601 J potential = W/t = 20601 J / (60 m /2m/s) = 686 Watts 40 9. A. P = W/t = F*d /t t = F*d/P t = d/v v = P/F v = 2 hundred W / (50kg * 9.80 one m/s^2) v = .407 m/s B. potential = 2 hundred W (The power = F*d = 2452 J) *********** Bhashkar components an impressive answer. you may learn some thing from his attack. yet interior the tip, he accidently substituted 323 N for 343 N. If he hadn't made that blunders, we the two would have had the same answer of 686 Watts (686 J/s). (We extensively utilized slightly distinctive values for g -- 9.81m/s^2 and 9.80 m/s^2 -- yet g varies counting on area)