00:01
This problem, we're going to talk about gravitational acceleration.
00:03
Remember that the gravitational acceleration g is equal to 9 .8 meters per second squared.
00:10
And if we set up a coordinate system such that the y axis is pointing upwards, then since g is pointing downwards towards the center of the earth, the position y of a particle subject to the gravitational acceleration is equal to y0, that is the initial position, plus the initial velocity v0 times the the time delta t minus g over two delta t squared and where the minus sign here appears because the gravitational acceleration points downwards so all we have in our problem is a basketball player that jumps from the ground to a height of 1 .1 meters and our goal is to find how long the player stays on the air.
01:05
Notice that we can draw a graph of the position of the player is a touch of time and it would look something like this.
01:15
We want to calculate this time here, calling in delta t.
01:21
Notice that delta t is equal to two times the time that i'm going to call t 1 1ā2.
01:28
So delta t is 2 times t 1 1 half.
01:34
Where 2 .2.
01:35
Is the time that it takes for the player to go from the ground to the maximum point in the trajectory or from the maximum point in the trajectory to the ground.
01:49
And the origin of my coordinate system is exactly on the ground...