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(C) Let the ball be thrown with velocity components $v_x$ and $v_y$. The bird moves horizontally with speed $u$. Since the ball touches the bird,the h

Vedclass · Accepted Answer

$2u \sqrt{\frac{2h}{g}}$

Vedclass · Answer

(C) Let the ball be thrown with velocity components $v_x$ and $v_y$. The bird moves horizontally with speed $u$. Since the ball touches the bird,the horizontal velocity of the ball must be equal to the speed of the bird,so $v_x = u$.The vertical displacement of the ball at the point of contact is $h$. Using the equation of motion $h = v_y t - \frac{1}{2}gt^2$ and the condition that the ball 'just touches' the bird (meaning the vertical velocity of the ball at that height is zero relative to the bird's path,or simply considering the peak of the trajectory),we have $h = \frac{v_y^2}{2g}$,which gives $v_y = \sqrt{2gh}$.The time taken to reach the height $h$ is $t = \frac{v_y}{g} = \sqrt{\frac{2h}{g}}$.The total time of flight for the projectile is $T = 2t = 2\sqrt{\frac{2h}{g}}$.The horizontal range $R$ is given by $R = v_x T = u 	imes 2\sqrt{\frac{2h}{g}} = 2u\sqrt{\frac{2h}{g}}$.

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