Effect of Gravity on Earth If a rock falls from a height of 20 meters on Earth, the height H( in

step 1 Here in this question, it is being said that there is a rock. I'll just draw a rough diagram ove

Effect of Gravity on Earth If a rock falls from a height of...

Key Concepts

Quadratic Functions

A quadratic function is a polynomial function of degree two, typically expressed in the form ax² + bx + c. These functions graph as parabolas and are used to model many natural phenomena. Their properties, such as the vertex, axis of symmetry, and roots, are crucial for understanding the behavior of systems that have accelerating motion or other quadratic relationships.

Free Fall Kinematics

Free fall kinematics deals with the motion of objects under the force of gravity, assuming negligible air resistance. In these problems, the distance fallen is proportional to the square of the time elapsed, which is reflected in the quadratic term of the motion equation. Understanding free fall helps in analyzing how objects accelerate towards the Earth.

Gravitational Acceleration

Gravitational acceleration is the constant acceleration due to gravity that objects experience when falling near the Earth's surface. This acceleration is denoted by g, which approximates 9.8 m/s², and is typically halved in the quadratic term when deriving displacement equations from acceleration formulas. This concept is fundamental in linking theoretical models to the tangible motion of falling objects.

Evaluating Functions

Evaluating functions involves substituting a specific value for the variable in an equation to find the corresponding output. In the context of physical motion, this process allows one to determine measurements like height or speed at a particular moment in time. It is an essential step in applying mathematical models to real-world situations.

Solving Equations

Solving equations involves finding the variable values that satisfy a given equation. In the context of quadratic functions and physical models, techniques such as factoring, completing the square, or using the quadratic formula are commonly employed. This process is crucial when determining key moments, like the time when an object reaches a specific height or impacts the ground.

Transcript

00:01 Here in this question, it is being said that there is a rock.

00:05 I'll just draw a rough diagram over here.

00:07 So this is the rock over here and it is falling on the ground.

00:13 So you can say earth and the height of this rock is 20 meters from the earth.

00:20 So this distance is 20 meters.

00:23 Fine.

00:24 Now, what is being said over here? you've been given an expression that expresses the height.

00:30 In meters after x seconds and the expression that has been given versus it's h is what h is the height after x seconds so h x is equal to 20 minus 4 .9 x square x is what x is the seconds okay the time so this expression expresses the height after x seconds now moving on to the question that is being asked see i hope the whole scenario is clear with all a few days rock that is falling onto the earth and the distance of the rock is 20 meters fine now we have to find out the height of the rock when x is equal to one seconds okay x is one and you have to find out the height or i can say you just have to find out edge of one see x is one no so that means just put x equal to one in the given expression so that will be 20 minus 4 .9 1 square isn't it? and one square is simply 1, so 4 .9.

01:35 And 20 minus 4 .9 gives you 15 .1 meters.

01:42 This is meter, no? the unit of the height is of the distance you can say is meters.

01:47 So the answer will be 15 .1 meters.

01:50 That means the height of the rock after one second is 15 .1.

01:58 Apart from that, you can also ask.

02:00 To find out the height when x is equal to 1 .1 second.

02:05 So now for that what you have to do is you have to find out edge of 1 .1.

02:10 That means see when x is equal to 1 .1 no you have to find out the height of the rock when the time is like after falling after falling after 1 .1 second.

02:23 So now i'm going to substitute x equal to 1 .1 in the given expression.

02:29 So like this.

02:31 Now 1 .1 square.

02:34 What is 1 .1 square? that is 1 .21.

02:38 Isn't it? now, simply 20 minus 4 .9 times 1 .21.

02:45 That will give you 5 .929.

02:49 And 20 minus 5 .929 is equal to 14 .071 meters.

02:56 So h of 1 .1 is equal to 14 .071 meters.

03:05 Now again, there are many parts to the, i mean, many subparts to the same part, you can say.

03:11 We've already found out the distance or the height when x is equal to one second as well as 1 .1 seconds.

03:18 Now you have to find out for x equal to 1 .2 second as well.

03:24 So simply h of 1 .2 again, just follow the same.

03:27 Same thing substitute x as 1 .2 in the given expression like this so 20 minus 4 .9 times 1 .2 square 1 .2 square is equal to 1 .44 fine and 20 minus 4 .9 times 1 .4 gives you 7 .056 now 20 minus 7 .056 is equal to 12 .944 meters.

04:02 So this is the required height.

04:05 Again, you have to find out the height when x is equal to 1 .3 seconds as well.

04:11 So same thing again, just the one difference that this time i'll substitute x as 1 .3 instead.

04:19 Now 20 minus 4 .9, 1 .3 whole square gives you 1 .69.

04:29 Is 4 .9 times 1 .69 now gives you 8 .281 and now just do the simple subtraction 20 minus 8 .281 gives you 11 .719 meters.

04:46 So this is the required height for all different 4 seconds you can say...

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