Question #98e74

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 #v_0# and hits the ground on its way down with the same velocity #v_0#.

So if it takes gravity a certain time to stop an object launched straight up with #v_0#, it will take gravity the exact same time to accelerate the object from rest to #v_0#.

Since you know that the total time of motion is equal to #"6.0 s"#, you can say that it's going to take #"3.0 s"# for the object to go from ground level to maximum height, which we'll call #h#.

This means that you can say

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

Here

• #v_"top"# is the velocity of the object at 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 #"0 m s"^(-1)#, which means that the above equation becomes

#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 #"3 s"# to get

#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 #color(orange)("(*)")# to get the value of #h#. Rearrange the equation to isolate #h#

#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")))#