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(B) The initial velocity components are $u_x = 3 	ext{ m s}^{-1}$ and $u_y = 4 	ext{ m s}^{-1}$.Using the equations of motion: $x = u_x t = 3t \Righ

Vedclass · Accepted Answer

$-5$

Vedclass · Answer

(B) The initial velocity components are $u_x = 3 	ext{ m s}^{-1}$ and $u_y = 4 	ext{ m s}^{-1}$.Using the equations of motion: $x = u_x t = 3t \Rightarrow t = \frac{x}{3}$.$y = u_y t - \frac{1}{2} g t^2 = 4t - \frac{1}{2} (10) t^2 = 4t - 5t^2$.Substituting $t = \frac{x}{3}$ into the equation for $y$:$y = 4 \left( \frac{x}{3} ight) - 5 \left( \frac{x}{3} ight)^2 = \frac{4x}{3} - \frac{5x^2}{9}$.To match the form $\frac{1}{9} [\beta x + \gamma x^2]$, we multiply and divide by $9$:$y = \frac{1}{9} [12x - 5x^2]$.Comparing this with the given form $\frac{1}{9} [\beta x + \gamma x^2]$, we get $\beta = 12$ and $\gamma = -5$.

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