A body is projected from the ground with a ve | vedclass

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(A) Given, initial velocity vector $v = (3 \hat{i} + 10 \hat{j}) 	ext{ m/s}$.Comparing with $v = v_x \hat{i} + v_y \hat{j}$, we get $v_x = 3 	ext{ m

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$5 	ext{ m}$ and $6 	ext{ m}$

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(A) Given, initial velocity vector $v = (3 \hat{i} + 10 \hat{j}) 	ext{ m/s}$.Comparing with $v = v_x \hat{i} + v_y \hat{j}$, we get $v_x = 3 	ext{ m/s}$ and $v_y = 10 	ext{ m/s}$.The maximum height $H$ attained by a projectile is given by the formula $H = \frac{v_y^2}{2g}$.Substituting the values: $H = \frac{10^2}{2 	imes 10} = \frac{100}{20} = 5 	ext{ m}$.The horizontal range $R$ is given by the formula $R = v_x 	imes T$, where $T = \frac{2v_y}{g}$ is the time of flight.Calculating time of flight: $T = \frac{2 	imes 10}{10} = 2 	ext{ s}$.Calculating range: $R = 3 	imes 2 = 6 	ext{ m}$.Thus, the maximum height is $5 	ext{ m}$ and the range is $6 	ext{ m}$.

$A$ body is projected from the ground with a velocity $v = (3 \hat{i} + 10 \hat{j}) \text{ m/s}$. The maximum height attained and the range of the body respectively are (given $g = 10 \text{ m/s}^2$):

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