A mass 1kg has kinetic energy 100J and it is thrown vertically upward. What is the maximum height it can attain?

Explanation:

There are a couple of different ways in which you can solve this and still get the same answer rounded. Since we are talking about KE, I will show you the way that utilizes KE and PE to get the total energy and then figure from that how high the object can go. First off, I'm going to use a mass of 1.0 kg for 2 significant digits. The total energy of a system is found in the equation

TE = PE + KE that says the total energy available to a system is equal to its kinetic energy plus its potential energy. Right off, we are given the KE value of 100 (Even though it's not accurate, I'm going to say that that number has 3 sig fig's, just because rounding to 1 sig fig is counterproductive). If the KE is 100, then

TE = 100 + 0 (the PE is 0 if the object is not moving, which it's not if someone is holding it and then throws it upwards). That's the total amount of energy available to that system and it cannot go up and it cannot go down, it can only change form. If the TE = 100, then we move on to the second part of the problemwhich is finding out how high i can go. The max height of the object indicates that the KE is 0 (the object at its max height isn't moving, even though it's only not moving for a nanosecond. If the object is not moving AND it's at its highest point, KE is 0 and PE is at a max).  That means that at this max height,

TE = 0 + PE and filling in the value for TE:

100 = PE and PE = mgh where m is mass, g is the pull of gravity, and h is the height (our unknown).

100 = (1.0)(9.8)(h) and

so, to 2 sig fig's,

h = 1.0 × 10¹ meters (or 10 meters)

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr