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(C) To calculate the maximum height attained, we consider only the vertical component of the motion.At the given height $h_1 = 5 	ext{ m}$, the verti

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$6.25$

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(C) To calculate the maximum height attained, we consider only the vertical component of the motion.At the given height $h_1 = 5 	ext{ m}$, the vertical component of velocity is $u_y = 5 	ext{ m s}^{-1}$.At the maximum height, the vertical component of velocity becomes $v_y = 0 	ext{ m s}^{-1}$.The acceleration due to gravity is $a_y = -10 	ext{ m s}^{-2}$.Using the kinematic equation $v_y^2 - u_y^2 = 2 a_y s$, where $s$ is the additional height gained:$0^2 - (5)^2 = 2(-10) s$$-25 = -20 s$$s = \frac{25}{20} = 1.25 	ext{ m}$The total maximum height reached by the ball is $H = h_1 + s = 5 	ext{ m} + 1.25 	ext{ m} = 6.25 	ext{ m}$.

$A$ ball is projected from the ground into the air. At a height of $5 \text{ m}$, its velocity is $\vec{v} = (5 \hat{i} + 5 \hat{j}) \text{ m s}^{-1}$. The maximum height reached by the ball is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$): (in $\text{ m}$)

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