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(B) Let the initial velocity of the body be $u$. At the maximum height $H$, the final velocity $v = 0$. The formula for maximum height is $H = \frac{u

Vedclass · Accepted Answer

$3.2$

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(B) Let the initial velocity of the body be $u$. At the maximum height $H$, the final velocity $v = 0$. The formula for maximum height is $H = \frac{u^2}{2g}$.At a height $h$, the velocity is $v_h = 8 \,m/s$. Using the equation of motion $v^2 = u^2 - 2gh$, we have $8^2 = u^2 - 2gh$, which implies $u^2 = 64 + 2gh$.However, the problem implies the body reaches a height $h$ where the speed is $8 \,m/s$. Assuming the body is projected from the ground, the maximum height $H$ is reached when the velocity becomes $0$. If we consider the motion from the point of height $h$ to the maximum height $H$, the initial velocity is $8 \,m/s$ and final velocity is $0$. The distance covered is $(H - h)$.Using $v^2 = u^2 - 2as$, we get $0^2 = 8^2 - 2g(H - h)$, so $64 = 20(H - h)$, which gives $(H - h) = 3.2 \,m$.If the question implies the body is at height $h$ and the speed is $8 \,m/s$, the additional height gained is $3.2 \,m$. Assuming the body was projected such that $h$ is the point where speed is $8 \,m/s$, the maximum height relative to that point is $3.2 \,m$.

$A$ body is projected vertically upwards. It reaches a point at a height $h$ above the point of projection after some time with a speed of $8 \,m/s$. The maximum height reached by the body is (Acceleration due to gravity $g = 10 \,m/s^2$). (in $m$)

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