Question

From the concept of directed dimension what is the formula for a range (R) of a cannon ball when it
is fired with vertical velocity component V and a horizontal velocity component V assuming it is
fired on a flat surface. [Range also depends upon acceleration due to gravity, g and k is numerical
constant)

Answer 1

Answer:

plz mark it the brainliest.. hope it helps you

Answer 2

Given,

Vertical and Horizontal Velocity Component = $V$

Acceleration due to gravity = $g$

Numerical Constant = $k$

Now, the horizontal component of the velocity is shown by $V_{x}$ and the vertical component by $V_{y}$.

The dimensional formula of the horizontal and vertical components can be given by,

$V_{x}$ = $L_{x} T^{-1}$

$V_{y}$ = $L_{y}T^{-1}$

The dimensional formula of Acceleration due to gravity can be given by,

$g = L_{y} T^{-2}$

The range can be given by,

$R = k [V_{x}]^{a} [V_{y}]^{b}[g]^c$

$L_{x} = [L_{x}T^{-1}]^a [L_{y}T^{-1}]^b [L_{y}T^{-2}]^c$

$L_{x} = [L_{x}^{a} L_{y}^{b-c} T^{-a-b-2c}]$

On comparing we get,

$-a-b-2c = 0$

$b = -c$

$b+2c = -1$

After solving the equations we get,

$a = 1, b = 1, c = -1$

Hence the range will be = $k[V_{x}]^a[V_{y}]^b[g]^c$

Range = $\frac{k [V_{x}] [V_{y}]}{g}$

#SPJ2