# Question #98e74

##### 1 Answer

#### Explanation:

The trick here is to realize that the time needed for the object to go from ground level to maximum height is **equal** to the time needed for the object to go back from maximum height to ground level.

That is the case because the object is moving **under gravity**, meaning that gravity slows it down while going up and accelerates it while coming down again.

In other words, the object is lanched with

So if it takes gravity a certain time to stop an object launched straight up with

Since you know that the *total time* of motion is equal to

This means that you can say

#v_"top"^2 = v_0^2 - 2 * g * h#

Here

#v_"top"# is the velocity of the objectat maximum height#v_0# is its initial velocity#g# is the gravitational acceleration, equal to#"9.81 m s"^(-2)#

But you know that at maximum height, the velocity of the object is equal to

#v_0^2 = 2 * g * h" " " "color(orange)("(*)")#

Now, you can determine the initial velocity of the object by using the fact that

#v_"top" = v_0 - g * t#

This is equivalent to

#v_0 = g * t#

Use the fact that the time needed to reach maximum height is equal to

#v_0 = "9.81 m s"^color(red)(cancel(color(black)(-2))) * 3color(red)(cancel(color(black)("s"))) = "29.43 m s"^(-1)#

Now all you have to do is plug this value into equation

#v_o^2 = 2 * g * h implies h = v_0^2/(2 * g)#

and plug in your value to find

#h = (29.43^2 "m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-2)))))/(2 * 9.81 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2))))) = "44.145 m"#

Rounded to two **sig figs**, the number of sig figs you have for the total time of motion, answer will be

#color(darkgreen)(ul(color(black)(h = "44 m")))#