A cricket bowler releases the ball in two different ways

(a) Giving it only horizontal velocity, and 

(b) Giving it horizontal velocity and a small downward velocity. 

The speed vs at the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation – horizontal velocity u_x= v_s

During projectile motion horizontal velocity remains unchanged

V_x=u_x=v_s

In vertical direction v_y²= u_y²+2gH

V_y= $\sqrt[]{2gH}$

Resultant speed of the ball at the bottom

V= $\sqrt[]{v_x^2+v_y^2}$

V= $\sqrt[]{v_s^2+2gH}$

b) when ball is given horizontal velocity and a small downward velocity

in horizontal direction V'_x=u_x=v_s

in vertical direction v_y'²= u_y²+2gH

resultant velocity of the ball at the bottom

v'= $\sqrt[]{v_s^2+u^2+2gH}$

 

<p><span lang="EN-IN">This is a Long Answer type Questions as classified in NCERT Exemplar</span></p><p><strong><span lang="EN-IN">Explanation –</span></strong><span lang="EN-IN"> horizontal velocity u_x= v_s</span></p><p><span lang="EN-IN">During projectile motion horizontal velocity remains unchanged </span></p><p><span lang="EN-IN">V_x=u_x=v_s</span></p><p><span lang="EN-IN">In vertical direction v_y²= u_y²+2gH</span></p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1751262523phpAzMUWq.jpeg" alt="" width="277" height="223" loading="lazy"></picture></div><div class="figure"><p><span lang="EN-IN">V_y= <span class="mathml" contenteditable="false"> <math> <mroot> <mrow> <mrow> <mn>2</mn> <mi>g</mi> <mi>H</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></span><span lang="EN-IN">&lt;!- [endif]-&gt;</span>&lt;!- [endif]-&gt;</p><p><span lang="EN-IN">Resultant speed of the ball at the bottom </span></p><p><span lang="EN-IN">V= <span class="mathml" contenteditable="false"> <math> <mroot> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>+</mo> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></span><span lang="EN-IN">&lt;!- [endif]-&gt;</span>&lt;!- [endif]-&gt;</p><p><span lang="EN-IN">V= <span class="mathml" contenteditable="false"> <math> <mroot> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>+</mo> <mn>2</mn> <mi>g</mi> <mi>H</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></span><span lang="EN-IN">&lt;!- [endif]-&gt;</span>&lt;!- [endif]-&gt;</p><p><span lang="EN-IN">b) when ball is given horizontal velocity and a small downward velocity</span></p><p><span lang="EN-IN">in horizontal direction V'_x=u_x=v_s</span></p><p><span lang="EN-IN">in vertical direction </span><span lang="EN-IN">v_y'²= u_y²+2gH</span></p><p><span lang="EN-IN">resultant velocity of the ball at the bottom </span></p><p><span lang="EN-IN">v'=<span class="mathml" contenteditable="false"> <math> <mroot> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>+</mo> <msup> <mrow> <mrow> <mi>u</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <mn>2</mn> <mi>g</mi> <mi>H</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></span><span lang="EN-IN">&lt;!- [endif]-&gt;</span>&lt;!- [endif]-&gt;</p></div></div><p>&nbsp;</p>

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