00:01
So in this question, we have the fact that the kangaroo is going to jump all the way over a 2 .5 meter object.
00:09
So if this is 2 .5 meters, the kangaroo is going to jump just over it and come back down.
00:17
Now, we only care about the vertical motion, so it doesn't matter how it moves in the x.
00:23
For part 8, we need to find the initial vertical speed.
00:46
And for the second part, we are going to find time spent in there.
01:00
So what is our equation for y? the equation of the motion in y is y of t equals to, and based on the equations we have, is a over 2, t squared, plus initial velocity or v0, t plus h, or why not for initial height.
01:37
Our initial height is going to be zero so we can just cross this out.
01:43
And now we need to find initial speed.
01:48
Well, the only fact that we have is that maximum height is 2 .5 meters.
01:53
When is the maximum height going to happen? in order to find it, we need to take the derivative of y.
02:01
And that would give us 80.
02:04
Plus v0.
02:07
And maximum happens when derivative is zero...