00:02
Hi, we have a helicopter at 9 .5 meter above the ground that descends with 4 .1 meter per second speed.
00:12
The y direction is positive direction is up forth.
00:15
So that means that the velocity of the helicopter is negative because it is moving in downwards direction.
00:22
Now the helicopter drops a package from the rest with respect to the helicopter.
00:28
That means at that instant the velocity of the package is also minus 4 .4.
00:34
We need to find what is the velocity of the package as it hits the ground, calculated with respect to the helicopter.
00:44
The first step here should be to find the velocity of the package as it hits the ground.
00:51
So we can calculate it using the formula v final is equal to v initial plus acceleration times the time, where we initial is the initial velocity of the package, which is minus 4 .10 meter per second.
01:03
It is moving downwards.
01:05
Acceleration value is minus g or minus 9 .8 meter per second square.
01:11
But we can see that we don't know the time.
01:13
The time can actually be found from another kinematic equation for the vertical direction.
01:20
Y final minus y initial, where y final is at zero at the ground, y initial is at 9 .5 meter height.
01:29
Vy initial again is the initial velocity of the package, minus 4 .4.
01:33
10 meter per second and acceleration is minus g so from here let's find the time to find the time from here we can see but y final minus y initially simply 0 minus 9 .5 meter is equal to minus 4 .10 meter per second times the time plus one half times minus 9 .8 meter per second square times the time times square.
02:10
Now we simplify it, let's drop the unit for a moment, units, so it is easier to check the equation.
02:18
And then from here we have 4 .9 t squared plus 4 .1 t minus 9 .5 is equal to 0.
02:32
This is quadratic equation.
02:33
We can find the time and the time from here is equal to minus 4 .1 plus we are only using the plus because minus gives us negative time which we can not use so plus 4 .1 square minus 4 times 4 .9 times minus 9 .5 this is our usual quadratic equation minus b plus minus under the square minus 4 a c divided to 2a which is 2 times 4 .9.
03:17
When we calculate here we get 1 .04 seconds.
03:23
This is the time after which the package hits the ground.
03:28
Now we can substitute it in our equation.
03:32
V -final is equal to v -initial plus acceleration times time...