00:01
Hi there, so for this problem, we are asked these two questions for part a and part b.
00:09
However, i'm not given the information, but i'm going to assume some values, and then you can change them accordingly to the values that you are given, okay? so this problem should be that a small rock is thrown vertically upward, so there is only a white component of the speed, no, an horizontally component of the speed.
00:30
And that will be equal to, well, let's assume that that is equal to 23 meters per second.
00:37
Okay, again, you can change that accordingly.
00:40
The equations are going to be the same.
00:42
From the edge of a roof of a 25 meters tall building.
00:46
So the height, the initial height for this, let's call that y -0, is equal to 25 meters.
00:55
Now, the rock doesn't hit the building on its way back down.
01:00
And lands in the street below.
01:03
Now, for part a of this problem, the question is about what is the speed of the rock just before it hits the street? so, in that case, first what we need to determine is the time that it takes to get to the ground, okay? so we know that the initial from kinematics, we know that the vertical position is equal to the initial position plus the initial speed times the time this minus 1 divided by 2 times the acceleration due to gravity times the time s squared.
01:53
Okay.
01:53
Now we want to calculate the speed when it hits the ground.
01:59
So we set this equal to zero because that is the altitude at the time level.
02:03
So that will be in here we substitute the values 25 for the initial.
02:09
Is an altitude plus the initial speed that is 23 times the time.
02:15
This minus 1 divided by 2.
02:17
The acceleration due to gravity that we know is 9 .8 meters per second square.
02:22
So that divided by 2 is just 4 .9...