A model rocket is launched straight upward with an initial...

A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 ${m} / {s}^{2}$ until its engines stop at an altitude of 150. m. (a) What can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

Key Concepts

Kinematic Equations

The kinematic equations describe the motion of an object experiencing constant acceleration. They are useful in solving problems involving displacement, velocity, time, and acceleration, such as determining the rocket's height, speed, and duration of various phases of its flight.

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is influenced only by gravity (neglecting air resistance) after its initial propulsion. This concept helps in predicting the trajectory, maximum height, and time in the air once the rocket's engines have stopped.

Phase-wise Analysis

Phase-wise analysis involves breaking down the motion of the rocket into distinct parts, such as the powered flight (while the engines are active) and the unpowered flight (when the only force acting is gravity). This approach simplifies the problem by allowing the use of different initial conditions and accelerations for each phase.

Acceleration Due to Gravity

Acceleration due to gravity is a constant acceleration that acts downward near the Earth's surface, typically valued at 9.8 m/s². This key concept is crucial in determining the unpowered flight behavior of the rocket, as it affects how high the rocket will coast and the time it takes to reach its peak after the engines cut off.

Transcript

00:01 Hi friends, this is the problem based on motion of the rocket under gravity.

00:26 Here it is given from the ground a rocket is launched with initial velocity of 50 meter per second with acceleration of 2 meter per second square at altitude of 150 meter above the ground engine is stopped that is acceleration of rocket become zero in the first part we have to describe motion of the rocket after its engine is engine stops let us hear the answer when its engine is stopped then rocket is moving freely under gravity.

02:20 Now second part we have to calculate maximum height attained by it at max.

02:37 Velocity of rocket at altitude of 150 meter b .y not is sculpto.

02:57 V0 plus that is 50 plus but time is not given i'm sorry for it you should apply v y square is called to v0 square plus 2 a r into y not so it is to be 50 plus 2 into 2 into 150 so on solving it, it is to be just a moment please 55 .78 meter per second.

04:07 So maximum height attained.

04:09 So at maximum height, final velocity becomes zero.

04:15 So applying the equation, v is equal to v0 square plus 2g h -mex.

04:25 This is 0 55 .78 to 9 .8 hmex.

04:43 So you will get so it is to be 158 .18.

05:07 Hence total height above to the ground y0 plus hmex, y0 is 1508.

05:17 158 .158.

05:20 So it is 200, sorry 308 .2 meter.

05:33 And c part we have to find the time taken, time taken to attain hmx t1, time taken to reach from ground to 150 meter with initial velocity...