00:01
Hello students, in the first problem the ball returns to its starting point in 4 seconds.
00:08
So, time taken to reach maximum height is 2 seconds and at maximum height final velocity is 0.
00:19
So, we can write the equation of motion k equals to u minus kp.
00:25
Therefore, 0 equals to u minus 1 .6 into 2.
00:31
So, from this we have u will be initial speed is 3 .2 meter per second.
00:41
Now, in the next problem initial speed of the truck u equals to 0 and acceleration is 5 meter per second square.
00:53
So, after 4 seconds speed is u plus at that is 0 plus 5 into 4 meter per second.
01:10
So, that is equals to 20 meter per second and distance traveled equals to ut plus half a t square which is equals to 0 plus half into 5 into 4 square and the unit of meter.
01:37
So, distance traveled is 40 meter.
01:44
Now, in problem 3 acceleration is v minus u by p.
01:53
So, that is 2 .7 minus 0 divided by 3 meter per second square.
02:02
So, that is equals to 0 .9 meter per second square.
02:10
So, distance moved in 6 second distance move in 6 second is equals to 2t plus half a t square that is u is 0.
02:27
So, half half into 0 .9 into 6 square in the unit of meter...