Question

A stone is thrown upwards from the ground with a velocity ‘u’
a) Obtain the maximum height attained by the stone?
b) Check the correctness of the equation obtained in (a) using the
method of dimensional analysis.

Answer 1

$\text{a)the stone moves with equal slowness:}$

$V = V_0-gt$

$V_0=u$

$V=u-gt$

$\text{the maximum lifting height occurs when the stone begins to move down}$

$V=0$

$u-gt=0$

$t=\frac{u}{g}$

$\text{height of the stone above the ground:}$

$h = V_0t-\frac{gt^2}{2}=ut-\frac{gt^2}{2}$

$h_{max } \text{ when }t=\frac{u}{g}$

$h_{max}=u*\frac{u}{g}-\frac{g}{2}*\frac{u^2}{g^2}=\frac{u^2}{2g}$

$\text{b) dimensional analysis:}$

$h_{max}=\frac{u^2}{2g}$

$h_{max}-[m]$

$u-[m/s]$

$g-[m/s^2]$

$\frac{u^2}{2g}-[\frac{m^2/s^2}{m/s^2}]=[m]$

$\text{the dimensions of the left and right sides }$

$\text{of the formula coincide}$

Answer:$h_{max}=\frac{u^2}{2g}$