question_answer 79)

                  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 \[{{\upsilon }_{s}}\] at the time of release is die 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.

Answer:

                                  In first case, \[\upsilon =\,\sqrt{\upsilon _{x}^{2}+\upsilon _{y}^{2}}=\,\sqrt{\upsilon _{s}^{2}+2gH}\]                 In second case, \[{{\upsilon }_{x}}={{\upsilon }_{x}}\]             and \[\upsilon _{y}^{2}\,-\,\upsilon _{\sigma y}^{2}\,\,=2\,gH\] or \[{{\upsilon }_{y}}=\,\sqrt{\upsilon _{\sigma y}^{2}+2\,gH}\]                 \[\therefore \]\[\upsilon =\,\sqrt{\upsilon _{x}^{2}+\,\upsilon _{y}^{2}}=\,\sqrt{\upsilon _{s}^{2}+\upsilon _{\sigma y}^{2}+2gH}\]                 Since \[{{\upsilon }_{oy}}\,\]is very small, so \[\upsilon \,=\,\sqrt{\upsilon _{s}^{2}+2gH}\] OR                 In case (a), Total initial energy at height H,                 \[{{E}_{i}}=\,\frac{1}{2}\,m\,\upsilon _{s}^{2}+\,mgH\]                 Total energy when it hits the ground                 \[=\frac{1}{2}\,m\,{{\upsilon }^{2}}\]                 According to law of conservation of energy                 \[\frac{1}{2}\,m\,{{\upsilon }^{2}}=\frac{1}{2}\,m\upsilon _{s}^{2}\,+\,mgH\]                 or \[\upsilon =\,\sqrt{\upsilon _{2}^{s}\,+2gH}\]                 In second case, total initial energy at height H                 or \[\upsilon =\,\sqrt{\upsilon _{s}^{2}+\,2gH}\]                 In second case, total initial energy at height H                 or \[\upsilon =\,\sqrt{\upsilon _{s}^{2}+\,2gh}\]                 \[=\frac{1}{2}m\,{{(\upsilon _{s}^{2}+{{\upsilon }_{oy}})}^{2}}+\,mgH\]                 Total final energy, when it hits the ground \[=\frac{1}{2}\,m\,{{\upsilon }^{2}}\]                 According to the law of conservation of energy.                 \[\frac{1}{2}\,m{{\upsilon }^{2}}=\frac{1}{2}\,m\,(\upsilon _{s}^{2}+\upsilon _{oy}^{2})+mgH\]                 \[\upsilon =\,\sqrt{\upsilon _{s}^{2}+\upsilon _{oy}^{2}+2gH}\]                 Since \[{{\upsilon }_{oy}}\] is small, so \[\upsilon =\,\sqrt{\upsilon _{s}^{2}+\,2gH}\]