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<h2 class="wp-block-heading">Key Concepts:</h2>



<ul class="wp-block-list"><li>&nbsp;Motion&nbsp;under&nbsp;gravity</li><li>&nbsp;Problems&nbsp;based&nbsp;on&nbsp;the&nbsp;motion&nbsp;under&nbsp;gravity</li><li>&nbsp;The&nbsp;graphical&nbsp;presentation&nbsp;of&nbsp;motion&nbsp;under&nbsp;gravity&nbsp;&nbsp;</li></ul>



<h3 class="wp-block-heading">Introduction:&nbsp;</h3>



The acceleration due to gravity ’g’ remains constant near the surface of the earth. So, all the practical examples of objects executing vertical motion under gravity can be taken as one-dimensional motion with constant acceleration. We can study motion under gravity using the kinematic equations.&nbsp;&nbsp;&nbsp;



We will study the two cases of motion under gravity.&nbsp;



Case 1: An object is released downward&nbsp;



Case 2: An object is thrown upward&nbsp;



<h2 class="wp-block-heading">Explanation:&nbsp;</h2>



<h3 class="wp-block-heading">Case 1: An object released downward&nbsp;</h3>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img loading="lazy" decoding="async" width="460" height="254" src="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3396.png" alt="An object released downwards" class="wp-image-9625" srcset="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3396.png 460w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3396-300x166.png 300w" sizes="auto, (max-width: 460px) 100vw, 460px" /></figure></div>


We know that the three kinematic equations to use when an object executes uniformly accelerated motion are:&nbsp;<div class="fillWidthInMob"><a loading='lazy' class="showAdInAllRes" data-toggle="modal" data-target=#homeLeadForm data-heading="Get Expert Guidance" href="#"><img class="img-fluid" alt="parallel" src=https://a.storyblok.com/f/128066/1100x230/00b28b4876/11-applying-for-collages_leaderboard-1100x230.jpg /></a></div>



v = u + at&nbsp;



s = ut +𝟏/𝟐 at2&nbsp;



v2 – u2 = 2as&nbsp;



For an object falling under gravity, acceleration = a = g,&nbsp;



Height = s = h&nbsp;<div class="fillWidthInMob"><a loading='lazy' class="showAdInAllRes" data-toggle="modal" data-target=#homeLeadForm data-heading="Get Expert Guidance" href="#"><img class="img-fluid" alt="parallel" src=https://a.storyblok.com/f/128066/1100x230/00b28b4876/11-applying-for-collages_leaderboard-1100x230.jpg /></a></div>



so the equations become:&nbsp;



v = u + gt&nbsp;



h = ut +𝟏/𝟐 gt2&nbsp;



v2 – u2 = 2gh&nbsp;



Suppose the object is released from rest, then the initial velocity = u = 0.&nbsp;



v = gt&nbsp;



h = 𝟏/𝟐gt2&nbsp;



v2 = 2gh&nbsp;



Suppose the object is released from rest, then the initial velocity = u = 0.&nbsp;



v = gt&nbsp;



h = 𝟏/𝟐gt2&nbsp;



v2 = 2gh&nbsp;



If the total time to reach the ground = t = T&nbsp;



h = 𝟏/𝟐 gt2 



Total time = T = √𝟐𝒈𝒉



Maximum speed with which it will reach the ground = vmaximum&nbsp;



v2 = 2gh ⟹ vmaximum = √𝟐𝒈𝒉



Problem 1: Two balls of masses 50gm and 100gm are dropped from the same height of 80m, find the time taken by the balls to reach the ground. (Ignore air resistance and take g = 10 m/s2).&nbsp;&nbsp;



Solution:&nbsp;



Since the equations of motion under gravity are independent of the mass of the object, the time taken by the ball to reach the ground will be the same and that is given by the equation:&nbsp;



T = √2h/g



=√2×80/10 = 4 sec&nbsp;



Thus, both the 50gm and 100gm balls will reach the ground after 4 seconds.&nbsp;



<h3 class="wp-block-heading">Case 2: An object thrown upward&nbsp;</h3>



We know that the three kinematic equations to use when an object executes uniformly accelerated motion are:&nbsp;


<div class="wp-block-image">
<figure class="aligncenter size-full"><img loading="lazy" decoding="async" width="295" height="340" src="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/vfd.png" alt="An object thrown upwards " class="wp-image-9662" srcset="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/vfd.png 295w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/vfd-260x300.png 260w" sizes="auto, (max-width: 295px) 100vw, 295px" /></figure></div>


&nbsp;v = u + at&nbsp;



s = ut + 𝟏/𝟐 at2



&nbsp;v2 = u2&nbsp; + 2as&nbsp;



For an object moving against gravity, acceleration = a = &#8211; g,&nbsp;



Height = s = h&nbsp;



So, the equations become:&nbsp;



v = u – gt&nbsp;



h = ut – 𝟏/𝟐 gt2



&nbsp;v2 = u2 – 2gh&nbsp;



When a ball is thrown upwards with velocity u, it moves against gravity. Gradually its velocity decreases and after reaching a certain height its velocity becomes zero.&nbsp;


<div class="wp-block-image">
<figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3394-1024x685.png" alt="A ball is thrown upward, then it falls back" class="wp-image-9623" width="620" height="414" srcset="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3394-1024x685.png 1024w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3394-300x201.png 300w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3394-768x513.png 768w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3394.png 1083w" sizes="auto, (max-width: 620px) 100vw, 620px" /></figure></div>


Initial velocity = u&nbsp;



At the maximum height attained h = H&nbsp;



Final velocity = v = 0&nbsp;



Time of ascent = ta&nbsp;



v = u + gt&nbsp;



0 = u – gta&nbsp;



Time of ascent = ta = u/g



Also, v2 – u2 = 2as&nbsp;



0 – u2 = –2gH&nbsp;



u = √2gH



Thus, time of ascent = ta = √2gH/g



Time of ascent = ta = √2H/g



When the ball falls, initial velocity = u = 0 and a = g&nbsp;&nbsp;



Time of descent = t = td&nbsp;



v = u + gtd&nbsp;



v = gtd&nbsp;



H = 1/2gtd2&nbsp;&nbsp;



&nbsp;td = √2H/g



Thus, we see that&nbsp;&nbsp;



Time of ascent = Time of descent&nbsp;



ta = td = √𝟐𝑯𝒈



Problem-2: A ball is thrown up with a speed of 40 m/s. Calculate the time of ascent and the maximum height it reaches. (Ignore air resistance and take g = 10 m/s2).&nbsp;&nbsp;



Solution:&nbsp;&nbsp;



Initial velocity = u = 40m/s&nbsp;



At the maximum height attained h = H&nbsp;



Final velocity = v = 0&nbsp;



Time of ascent = ta&nbsp;



v = u + gt&nbsp;



0 = u – gta&nbsp;



Time of ascent = ta = u/g = 40/10 = 4 s&nbsp;



Also, v2 – u2 = 2as&nbsp;



0 – u2 = –2gH&nbsp;



–402 = –2 × 10 × H&nbsp;



1600 = 20H&nbsp;



H = 80m, the ball will reach the maximum height of 80m.&nbsp;



<h3 class="wp-block-heading">Graphical presentation of motion under gravity:&nbsp;</h3>



<h4 class="wp-block-heading">Case 1:&nbsp;Graphs for a body released downward:&nbsp;</h4>



Here, the graphs show the motion of a ball influenced by gravity alone. The ball is released from a height of 20 meters above the ground. The position is marked at 0.5 s intervals, the ball’s position, velocity, and acceleration are shown as a function of time.&nbsp;



Table 1: For a body released downwards&nbsp;


<div class="wp-block-image">
<figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3393-1024x776.png" alt="" class="wp-image-9622" width="534" height="404" srcset="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3393-1024x776.png 1024w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3393-300x227.png 300w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3393-768x582.png 768w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3393-1536x1164.png 1536w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3393.png 1600w" sizes="auto, (max-width: 534px) 100vw, 534px" /></figure></div>

<div class="wp-block-image">
<figure class="aligncenter size-full"><img loading="lazy" decoding="async" width="369" height="371" src="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3391.png" alt="Graphs for a body released downwards " class="wp-image-9619" srcset="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3391.png 369w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3391-298x300.png 298w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3391-150x150.png 150w" sizes="auto, (max-width: 369px) 100vw, 369px" /></figure></div>


<h4 class="wp-block-heading">Case 2: Graphs for a body thrown upwards:&nbsp;</h4>



Here the graphs show the motion of a ball influenced by gravity alone. The ball is launched straight up from the ground with an initial speed of 20m/s. The position is marked at 0.5 s intervals, the ball’s position, velocity, and acceleration are shown as a function of time.&nbsp;


<div class="wp-block-image">
<figure class="aligncenter size-full"><img loading="lazy" decoding="async" width="380" height="379" src="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3392.png" alt="Graphs for a body thrown upwards" class="wp-image-9621" srcset="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3392.png 380w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3392-300x300.png 300w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3392-150x150.png 150w" sizes="auto, (max-width: 380px) 100vw, 380px" /></figure></div>


Table 2: For a body thrown upwards&nbsp;


<div class="wp-block-image">
<figure class="aligncenter size-large is-resized"><img loading="lazy" decoding="async" src="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3395-1024x890.png" alt="" class="wp-image-9624" width="505" height="439" srcset="https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3395-1024x890.png 1024w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3395-300x261.png 300w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3395-768x667.png 768w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3395-1536x1334.png 1536w, https://www.turito.com/learn-internal/wp-content/uploads/2022/08/image-3395.png 1600w" sizes="auto, (max-width: 505px) 100vw, 505px" /></figure></div>


Numerical 1: A ball of mass 200gm is thrown vertically up from the ground. It reaches a maximum height of 20m in 10s. Find the initial velocity of the ball.&nbsp;



<ol class="wp-block-list"><li>20m/s</li><li>10 m/s</li><li>15 m/s</li><li>12 m/s&nbsp;</li></ol>



The initial velocity of the ball = u&nbsp;&nbsp;



Final velocity of the ball = v = 0 m/s&nbsp;



Acceleration of the ball = a = g = –10 m/s2 (negative sign because it is moving against gravity)&nbsp;



Maximum height attended = h = 20m&nbsp;



Using equation v2 – u2 = 2gh&nbsp;



0 – u2 = 2 × (–10) × 20&nbsp;



–u2 = –400&nbsp;



u = 20 m/s&nbsp;



Thus, the initial velocity of the ball = 20 m/s&nbsp;



Numerical 2: A ball is thrown vertically down. Find the time taken by the ball to reach the ground from a height of 500m.&nbsp;



<ol class="wp-block-list"><li>20s</li><li>10s</li><li>5s </li><li>5s&nbsp;</li></ol>



When an object is just released from a certain height the initial velocity of the object = u = 0&nbsp;



The final velocity of the stone = v&nbsp;&nbsp;



Acceleration of the stone = a = g = 10 m/s2 (positive sign because it is moving in the direction of gravity)&nbsp;



Height = h = 500m&nbsp;



Using equation h = ut + ½ gt2&nbsp;



500 = 0 × t + ½ × 10 × t2&nbsp;



500 = 0 + 5t2&nbsp;



100 = t2&nbsp;



Time = t = 10 m/s&nbsp;



<h2 class="wp-block-heading">Summary:</h2>



<ul class="wp-block-list"><li>The acceleration due to gravity ‘g’ remains constant near the surface of the earth. So, all the practical examples of objects executing vertical motion under gravity can be taken as one-dimensional motion with constant acceleration.</li><li>For an object to fall under gravity:</li></ul>



Acceleration = a = g, Height = s = h, initial velocity =u = 0



v = gt, h = gt2, v = 2gh2



<ul class="wp-block-list"><li>If the total time to reach the ground = t = T</li></ul>



h = 1/2gt2 



<ul class="wp-block-list"><li>For an object moving against gravity</li></ul>



Acceleration = a = -g, Height = s = h



so the equations become: 



v= u &#8211; gt 



h = ut &#8211; 1/2gt2



v2= u2-2gh



Time of ascent = Time of descent



ta = td = √2H/g, Total time = T = √2gh 



Maximum speed with which it will reach the ground = vmaximum



v2 = 2gh = v maximum = √2gh