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(D) For a ball thrown vertically upwards with initial velocity $u$,the displacement $h$ at time $t$ is given by the kinematic equation:$h = u t - \fra

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$h / t$ versus $t$

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(D) For a ball thrown vertically upwards with initial velocity $u$,the displacement $h$ at time $t$ is given by the kinematic equation:$h = u t - \frac{1}{2} g t^2$Dividing both sides by $t$ (for $t 
eq 0$):$\frac{h}{t} = u - \frac{1}{2} g t$Rearranging this into the linear equation form $y = m x + c$:$\frac{h}{t} = (-\frac{g}{2}) t + u$Comparing this with $y = m x + c$,where $y = \frac{h}{t}$,$x = t$,the slope $m = -\frac{g}{2}$,and the intercept $c = u$.Thus,plotting a graph of $\frac{h}{t}$ versus $t$ yields a straight line.The value of the acceleration due to gravity $g$ can be obtained by multiplying the magnitude of the slope of this line by $2$.

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