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Dimensional Analysis

The vertical ...

Question

The vertical height $y$ and horizontal distance $x$ of a projectile on a certain planet are given by $x = (3 t) m, y = (4 t) m$ where $t$ is in seconds. Find the speed of projection $(i n m / s)$ ?

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Solution

$v_{x} = \frac{d (3 t)}{d t} = 3$

$v_{y} = \frac{d (4 t - {6 t}^{2})}{d t} = 4 - 0 = 4$

$v = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}}$

$= \sqrt{3^{2} + 4^{2}}$

$= \sqrt{25}$

$= 5 \frac{m}{s}$

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