0:00
High.
00:02
In the given problem, mass of the rocket is given as m is equal to 20 kilogram.
00:15
Its height to be covered before reaching to the ground is h is equal to 80 meter and its speed, initially speed of the rocket.
00:28
When its engine starts exerting a force in upward direction, is given as 30 meter per second and that is in downward direction as the rocket has to come to stop just before touching the ground so its final speed is given as zero now net force acting on the rocket will be the force due to engine acting vertically upward minus its weight m g acting vertically downward and using newton the second law of motion that can be put equal to the product of mass of the rocket with its acceleration.
01:12
So expression for its acceleration will become f minus m g divided by m where we have to find this constant force being exerted by the engine on the rocket.
01:28
So using third equation of motion here, which says vf square is equal to v i square plus two as.
01:49
For vf, this is 0.
01:51
For v i, this was 30 meter per second to the whole square, plus 2.
01:59
For a, this is f minus m g divided by m into for s.
02:06
This is h, means 80 meter.
02:12
So it can be written as 0 is equal to 900...