00:01
Here in this question the student is moving through a water slide.
00:07
So let us represent the water slide.
00:10
This is the water slide and here this is the horizontal distance traveled.
00:17
This distance is given in the question which is 3 .2 meter.
00:22
This is the student and here there is a velocity v1.
00:27
Here in the first part we need to calculate the, the airborne time of the student here.
00:36
So airborne time is to be calculated.
00:39
So here we need to take the values according to the y direction.
00:46
So here in the y direction, the velocity, initial velocity v1 is equal to 0 meter per second because here while moving down in y direction the initial velocity will be 0 and the final velocity v2 is unknown.
01:03
Here the acceleration a in the y direction can be taken at acceleration due to gravity which is 9 .8 meter per second square and the distance travel is given which is 3 .2 meter.
01:18
So here delta t is unknown.
01:21
So here we need to use the formula.
01:24
S is equal to ut plus half 80 square, the relationship between displacement and time.
01:30
So this relationship is to be used here.
01:33
So here, so here displacement is represented as delta d is equal to u means initial velocity v1 into delta t plus half into acceleration a into delta t square here we know that initial velocity v1 is equal to zero so this term becomes zero so here delta d is equal to half into acceleration 9 .8 meter per second square into delta t square here delta d is known so here delta t will be equal to the equation is square of 2 window delta d which is the distance 3 .2 meter divided by 9 .8.
02:11
Upon solving this we will get the time taken to be 0 .81 seconds so this will be the airborne time.
02:20
So this is part a answer.
02:25
Now we are moving on to the next part part b.
02:29
In part b we to calculate the horizontal displacement.
02:34
So since the horizontal displacement is to be calculated, we have to consider the x direction here.
02:42
So the values corresponding to x direction is to be taken here.
02:48
Here in the x direction, we can see that here the initial velocity is given in the question which is 4 .2 meter per second and the final velocity will be also the same because here in the x direction, there is no change...