The velocity of a body which hasfallen freely under gravity varies as gᵖhʷ ,where g is the acceleration due to gravity at the place and h is the height through which the body has fallen. Determine the values of p and w..

The correct option is p=1/2 and q=1/2 (where q corresponds to the exponent w of height h in the question statement).

To find the values of p and w (referred to as q in the options), we can use the method of dimensional analysis.

Let the velocity of the body be represented by $v$. According to the problem statement, the velocity varies as $g^p{h}^w$. We can write this relation as an equation with a dimensionless constant $k$:
$v=k{g}^p{h}^w$

Let us write down the dimensions of each physical quantity involved:
1. Dimensions of velocity $v$ (distance/time) are: $\left[v\right]=\left[L{T}^{-1}\right]$
2. Dimensions of acceleration due to gravity $g$ (acceleration) are: $\left[g\right]=\left[L{T}^{-2}\right]$
3. Dimensions of height $h$ (length) are: $\left[h\right]=\left[L\right]$
4. The constant $k$ is a dimensionless quantity, so it has no dimensions.

Now, substituting the dimensions of all quantities into our equation:
$\left[L{T}^{-1}\right]={\left[L{T}^{-2}\right]}^p{\left[L\right]}^w$

Simplifying the right-hand side of the equation:
$\left[L^1{T}^{-1}\right]=\left[L^p{T}^{-2p}\right]·\left[L^w\right]$
$\left[L^1{T}^{-1}\right]=\left[L^{p+w}{T}^{-2p}\right]$

For the equation to be dimensionally correct, the exponents of the corresponding dimensions on both sides must be equal.

Equating the power of time $T$ on both sides:
$-2p=-1$
$p=1/2$

Equating the power of length $L$ on both sides:
$p+w=1$

Substitute the value of $p=1/2$ into the equation:
$1/2+w=1$
$w=1-1/2$
$w=1/2$

By mapping the exponent $w$ from the question to $q$ in the options, we find:
$p=1/2$ and $q=1/2$ (since $q=w$).