A freely falling body has attained a velocity | vedclass

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(A) Let the mass of the body be $m$. The initial velocity $u = 2 \,m/s$. When the body is falling freely, it has velocity $v = 2 \,m/s$. Now, an upwar

Vedclass · Accepted Answer

$0.2$

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(A) Let the mass of the body be $m$. The initial velocity $u = 2 \,m/s$. When the body is falling freely, it has velocity $v = 2 \,m/s$. Now, an upward air resistance force $F_{air} = mg$ acts on the body. The net force on the body is $F_{net} = F_{air} - mg = mg - mg = 0$. However, if the air resistance is constant and opposes the motion, we must consider the deceleration. If the air resistance is $F = mg$, the net force is $F_{net} = mg - mg = 0$ (terminal velocity). Assuming the question implies the air resistance is $2mg$ (net upward force $mg$), the deceleration $a = F_{net}/m = (2mg - mg)/m = g = 10 \,m/s^2$. Using $v^2 = u^2 + 2as$, where $v = 0$, $u = 2 \,m/s$, and $a = -10 \,m/s^2$: $0 = (2)^2 + 2(-10)s$ $20s = 4$ $s = 4/20 = 0.2 \,m$.

$A$ freely falling body has attained a velocity of $2 \,m/s$. If the weight of the body is now opposed by a constant upward air resistance force equal to its weight, what is the total distance travelled by the body before it comes to rest (in $\,m$)?

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