If an object is thrown straight up and air resistance is negligible, then its speed when it returns to the starting point is the same as when it was released. If air resistance were not negligible, how would its speed upon retum compare with its initial speed? How would the maximum height to which it rises be affected?

A little later, you will be studying energy. One of the things conserved in a system is energy. You cannot get more energy out of a system than you have put in.

That fact explains how the speed at the end of the object's travels can be the same as the beginning. All energy must be accounted for.

The second part is not so easily proved or even shown easily. It can be indicated, though.

Suppose vi = 40 m/s

Suppose vf = 0 m/s In other words, you throw a rock up and you stop timing it when it reaches its maximum height.

a = - 9.81 Gravity goes in the opposite direction to vi

vf = vi^2 + 2*a*d

0 = 1600 - 2*9.81*d

-1600 = -19.62*d

1600 / 19.62 = d

d = 81.55 m.

Now leave everything else alone and double vi

0 = 80^2 - 2*9.81 d

-6400 = -19.62 * d

d = 326.2

Give or take a bit, the height = roughly 4 times what it was if the initial speed is doubled. 326.2 / 81.55 = 4