A ball is dropped from a height of 49,m. The | vedclass

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(D) The ball is dropped with an initial velocity $u = 0$. It is subjected to two constant accelerations: vertical acceleration due to gravity $(g)$ an

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All the above.

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(D) The ball is dropped with an initial velocity $u = 0$. It is subjected to two constant accelerations: vertical acceleration due to gravity $(g)$ and horizontal acceleration due to wind $(a)$.Since the initial velocity is zero,the net acceleration vector $\vec{a}_{net} = \vec{g} + \vec{a}$ is constant and directed at an angle. The ball will move in a straight line along the direction of the net acceleration vector $\vec{a}_{net}$. Thus,statement $(a)$ is correct.Since the ball moves in a straight line at an angle to the vertical,the displacement is the hypotenuse of a right-angled triangle where the vertical height is one side. Therefore,the actual distance (displacement) travelled is greater than the vertical height of $49\,m$. Thus,statement $(b)$ is correct.The time taken to reach the ground depends only on the vertical motion. Using the equation of motion $h = \frac{1}{2}gt^2$,we get:$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 49}{9.8}} = \sqrt{10} \approx 3.16\,s$. Thus,statement $(c)$ is correct.Since all statements are correct,the correct option is $(d)$.

$A$ ball is dropped from a height of $49\,m$. The wind is blowing horizontally. Due to wind,a constant horizontal acceleration is provided to the ball. Choose the correct statement$(s)$. (Take $g = 9.8\,m/s^2$)

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