A projectile of mass M explodes at thee highst point of its trajectory when it hase vlocity . The horizontal distance travelede btween launch and explosion is x_0 . Two fragments are produced with initiale velocitis parallel to the ground. They thenfollow their trajectories until they hitt he ground. The fragment of mass m_1 retuns exactly to the launch point of thei orginal projectile (of mass M) while thee othr fragment of mass m_2 hits the grounda t a distance D from this point. Disregardn iteraction with air and assume that massa ws conserved in the explosion (m_1 +m_2 =M) Determine the magnitude of the velocity of fragment 2 just before it hits theground. (a) ((gx_0 )/v) (b)(√((25)/9))v (c) (√(((25)/9)v^2 +(((gx_0 )/5))2)) (d)(√((5/3)x_0 v^2 +(((gx


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Question Number 190793 by 2kdw last updated on 11/Apr/23

$A projectile of mass M explodes at thee$ $highst point of its trajectory when it hase$ $vlocity . The horizontal distance travelede$ $btween launch and explosion is x_{0} . Two$ $fragments are produced with initiale$ $velocitis parallel to the ground . They$ $thenfollow their trajectories until they hitt$ $he ground . The fragment of mass m_{1} retuns exactly to the launch point of thei$ $orginal projectile (of mass M) while thee$ $othr fragment of mass m_{2} hits the grounda$ $t a distance D from this point . Disregardn$ $iteraction with air and assume that massa$ $ws conserved in the explosion (m_{1} + m_{2} = M) Determine the magnitude of the$ $velocity of fragment 2 just before it hits theground .$ $(a) \frac{g x_{0}}{v}$ $(b) \sqrt{\frac{25}{9}} v$ $(c) \sqrt{\frac{25}{9} v^{2} + (\frac{{g x}_{0}}{5}) 2}$ $(d) \sqrt{\frac{5}{3} x_{0} v^{2} + (\frac{{g x}_{0}}{v}) 2}$
Commented by mr W last updated on 12/Apr/23

$t e r r i b l y m a n y t y p o s!$ $d u e t o t h e m a n y t y p o s a l l a n s w e r s$ $g i v e n a r e w r o n g!$ $p l e a s e c h e c k y o u r t y p o s a n d f i x t h e m$ $a t f i r s t!$ $a n d p l e a s e m a k e p r o p e r l i n e b r e a k s!$
	Answered by mr W last updated on 12/Apr/23

	Commented by mr W last updated on 12/Apr/23

	$v_{1} = v$ $t = \frac{x_{0}}{v} = \frac{D}{v_{2}}$ $\Rightarrow v_{2} = \frac{D v}{x_{0}}$ $h = \frac{{g t}^{2}}{2} = \frac{g}{2} {(\frac{x_{0}}{v})}^{2}$ $v_{3}^{2} = v_{2}^{2} + 2 g h = {(\frac{D v}{x_{0}})}^{2} + {(\frac{{g x}_{0}}{v})}^{2}$ $\Rightarrow v_{3} = \sqrt{{(\frac{D v}{x_{0}})}^{2} + {(\frac{{g x}_{0}}{v})}^{2}}$
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