A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration o

step 1 We take positive y to be upward. Now there are two periods of constant acceleration. One is a y

A rocket blasts off vertically from rest on the launch pad...

Key Concepts

Piecewise Motion Analysis

This concept involves breaking down a motion problem into distinct segments, each with its own conditions or accelerations. When the forces acting on an object change at a certain point (for example, an engine failure causing a transition from powered flight to free-fall), analyzing the motion separately for each segment and then connecting the solutions using boundary conditions is crucial.

Kinematic Equations

Kinematic equations describe the relationships among displacement, velocity, acceleration, and time for objects moving with constant acceleration. They are foundational for analyzing motion in one dimension and are used to determine unknown quantities when some variables are given, making them essential in solving problems involving uniformly accelerated motion.

Free Fall Under Gravity

Free fall refers to the motion of an object under the influence of gravity alone, where the only acceleration is the constant acceleration due to gravity. This concept is important for understanding how objects move after power or thrust is removed, and it involves using a specific gravitational acceleration value (approximately 9.8 m/s² downward) in kinematic equations.

Maximum Height Condition

At the maximum height in vertical motion, the instantaneous vertical velocity of the object becomes zero, even though acceleration (typically due to gravity) continues to act. This condition is used as a boundary point to determine further kinematic parameters and is a critical concept when analyzing projectile or upward moving objects.

Time of Flight Calculation

Time of flight involves calculating the total duration of an object's journey, from the start of its motion until it returns to its initial level. In problems that involve multiple phases of motion, such as an initial powered ascent followed by free-fall descent, determining the respective durations of each phase and summing them is essential for evaluating the full trajectory.

Transcript

00:01 We take positive y to be upward.

00:05 Now there are two periods of constant acceleration.

00:10 One is a y to be 2 .5 meter per second square while the engines fire and then the acceleration will be negative 9 .8 meter per second square when the engines shut off.

00:33 And both of these are accelerations are constant during these times.

00:40 So we can use constant acceleration equations within each period.

00:45 So basically during this period and this period.

00:49 Now for the first part, so we find the speed and height at the end of the first 20 seconds.

00:57 So t is equal to 20 seconds.

01:00 And we will use ay to be 2 .5 meter per second square.

01:08 And initial velocity is zero and initial height is also zero.

01:17 Now to find the final velocity we will use this equation since we have all the parameters all the rest of the parameters to solve for vy so we directly plug them into this equation and we get this velocity to be 50 meter per second.

01:35 Now for height we will use this equation over here because for the same reason we have all the parameters needed to solve this equation and we get this value of height or why to be 500 meter next consider the motion from this point to the maximum height so for that we have why not to be 500 meter because we are starting from this point now so now y0 will be 500 meter v0y will be now this value, 15 meter per second, and at maximum height, velocity is 0, so vy is 0 now.

02:29 And obviously acceleration is now negative 9 .8 meter per second square.

02:39 Now we solve these equations to find the maximum height which is y.

02:47 So let's move to the next page for that so we use this equation now since we don't have they have no idea about the time taken to travel this distance and we need to solve for height so this equation is perfect to solve for from this we can write y minus y not to be equal to this value and we solve this for the parameters we have for the for the second motion and this gives us 128 meter this means that y will be equal to y0 plus 128 meter and why not is as you can see here is 500 meter so why will be 628 meter now the duration of this part of the motion which is time could be found from this is an equation the same that we used here and from here we can find an expression for time we solve it for the values we have and this time comes out comes out to be 5 .10 seconds with the second part now at the highest point we know that velocity is zero because it comes to momentarily it comes to arrest and then falls down again but at that instant we why is zero and a y is obviously 9 .8 meter per seconds downward and if you are taking positive y axis to be upward then you can write a y to be negative 9 point meter per second square as we did in this case...

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