The equation of motion of a stone,thrown vert | vedclass

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(C) The equation of motion is given by $s = ut - 6.3t^2$.To find the velocity $v$,we differentiate $s$ with respect to $t$:$v = \frac{ds}{dt} = \frac{

Vedclass · Accepted Answer

$37.8$

Vedclass · Answer

(C) The equation of motion is given by $s = ut - 6.3t^2$.To find the velocity $v$,we differentiate $s$ with respect to $t$:$v = \frac{ds}{dt} = \frac{d}{dt}(ut - 6.3t^2) = u - 12.6t$.At the maximum height,the velocity of the stone is zero:$v = 0$.Given that the stone reaches its maximum height at $t = 3$ $sec$,we substitute these values into the velocity equation:$0 = u - 12.6(3)$.$0 = u - 37.8$.$u = 37.8$ $cm/sec$.

The equation of motion of a stone,thrown vertically upwards is $s = ut - 6.3t^2$,where the units of $s$ and $t$ are $cm$ and $sec$. If the stone reaches its maximum height in $3$ $sec$,then $u =$ ......... $cm/sec$.

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