00:01
Hi, here in this given problem, mass of the rocket that is given as 7 ,800 kilograms, its upward acceleration, that is 2 .30 meter per second square, and height at which the engine fails, that is, h is equal, to 575 meter means if we consider this horizontal line to be the launch pad then the rocket initial motion of the rocket is accelerated suppose from a to b it is accelerated motion then the engine fails so now it will be retarded motion under acceleration to gravity so here it is retarded then the velocity will become zero for an instance and then the rocket will start falling down like this and finally we'll crash to the launch pad again so in the first part of the problem we have to find maximum height of the rocket above the launch pad so that is having two parts, first part h which is given to us as 535 meter and the second part h dash which we have to find.
02:13
So to find this height, first of all we will find speed of the rocket, velocity of the rocket achieved at point a.
02:22
So, for accelerated motion of the rocket, vf square, final speed at point b, using third equation vf squared equals to vi square at the ground, plus 2ah.
02:53
Here it is 0 as the rocket starts moving up from rest, plus 2 times of 2 .30 into 535.
03:05
So this vf will be given by squireoto 2 ,645 means this is 51 .4 meter per second.
03:19
Now the engine fails as the engine fails here.
03:31
So the forward motion will be retarded motion...