In the first stage of a two-stage rocket, the rocket is...

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of $3.50 \mathrm{~m} / \mathrm{s}^{2}$ upward. At $25.0 \mathrm{~s}$ after launch, the second stage fires for $10.0 \mathrm{~s}$, which boosts the rocket's velocity to $132.5 \mathrm{~m} / \mathrm{s}$ upward at $35.0 \mathrm{~s}$ after launch. This firing uses up all of the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Ignore air resistance. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the end of the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stagetwo rocket be moving just as it reaches the launch pad?

Key Concepts

Uniform Acceleration

This concept refers to motion in which the acceleration is constant, meaning that the velocity changes at a fixed rate over time. It is fundamental in solving problems where an object’s displacement, velocity, or time is determined using equations such as s = ut + (1/2)at² and v = u + at.

Kinematic Equations

Kinematic equations provide the mathematical relationships among displacement, initial velocity, final velocity, acceleration, and time. They are essential tools for analyzing the motion of objects under uniform acceleration and for breaking down multi-phase motions (such as powered flight versus free-fall) into solvable segments.

Projectile Motion

Projectile motion describes the behavior of an object that is launched into the air and moves under the influence of gravity alone after fuel or other propelling forces are exhausted. Analyzing projectile motion involves determining the maximum height, time of flight, and final impact velocity, taking into account the constant acceleration due to gravity.

Multi-Stage Dynamics

In problems involving a multi-stage rocket, the dynamics are divided into separate phases with different forces acting on the rocket. Each stage, such as the powered flight stage and the free-fall phase, is analyzed separately using appropriate initial conditions and acceleration values until the entire motion can be understood as a composite of these phases.

Transcript

00:01 First, let us explain the relation for distance travel by a rocket in the first stage, that is s1, is equal to u1 t1 plus half a1t1 square.

00:20 Now we have u1 is 0 meter per second, time is 25 seconds plus half.

00:30 We have acceleration is 3 meter per second.

00:34 3 .5 meter per second square and t is 25 seconds so s1 is 1093 .8 meter there's a distance travel by the rocket traveled at first stages 103 no sorry 9 3 .8 meters now speed of the rocket at the first state so that is v1 is equal to u1 plus a1 d1 we have u0 plus a is 3 .5 and time is 25 seconds that is 87 .5 meter per second therefore speed of the rocket at the end of the first stage end of first stages 87 .5 meter per second now for the second stage u1 is sorry speed is v2 is equal to u2 plus a 2 t2 if so let us use view 1 instead of u2 so v2 is v1 plus a 2 t 2 so a 2 is v2 minus v1 upon t 2 so we have um v2 is 135 135 132 0 5 minus v1 is 87 .5 upon t 2 is 10 seconds that is 4.

03:03 5 meter per second square so acceleration at second stage is 4 .5 now let's see the distance travels that is s2 is equal to v1 t2 plus half a 2 t2 square that is equal to 87 .5 t2 is a 2 is 4 .5 meter per second square plus 10 second square that is 1100 meter therefore distance traveled by the rocket in the second stages 1100 meter now consider after completion of the second stage of the rocket the only excretion takes place is the acceleration due to gravity so now express the relation for speed at maximum height that the rocket travels the maximum height that the rocket traveled v3 square is equal to v2 square plus 2 as 3 so we have s 3 is equal to s 3 is equal to v2 so we have v2 is 132 .5 minus v3 is 0 upon 2 into g is 9 .8.

05:05 It is 895 .7 meter...

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