00:01
In this question, its rocket at ground level is fired straight up from rest with a net upward acceleration of 25 meter per second.
00:09
So this is from, let's say this is the point a from where it starts to move up at 25 meter per second square for 12 seconds.
00:21
Let's say at b.
00:22
And at b, the engine turns off and the rocket continues to post upward with insignificant air resistance definitely.
00:28
Because then from b to c, it will move under the force of gravity, which is 9 .8 meter per second square downwards, and it will stop at c.
00:38
And to find a maximum elevation.
00:41
So first, we need to find this distance, let's call it h1.
00:45
Then we have to find this distance.
00:46
Let's say h2.
00:48
All right.
00:49
So i think it starts from rest.
00:51
So from a to b, i'm going to use the second equation of motion that h1 is.
00:59
Equal to v0 t plus a t square initial speed is zero acceleration is 25 and the time is 12 second so that's 12 square so that's where the calculator comes in is 12 square times 25 over 2 which is coming as 1 ,800 meters we are also supposed to find the speed at b because then only we can find the distance h2...