The equation of trajectory of a projectile is . If we assume g $$=$$ $$10$$ ms–$$2$$ the range of projectile (in$$ $$meters) is

When the particle again strikes the ground, y $$=$$ $$0$$⇒ x $$=$$ $$0$$ or $$18$$ m Range $$=$$ $$18$$ m

Projection Under uniform Acceleration

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Question

The equation of trajectory of a projectile is $\large y = 10x - \left( {\frac{5}{9}} \right){x^2}$ . If we assume g = 10 ms^–2 the range of projectile (in meters) is

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Solution

$\large y = 10x - \left( {\frac{5}{9}} \right){x^2}$
When the particle again strikes the ground, y = 0
$\therefore 10x - \left( {\frac{5}{9}} \right){x^2}=0$
⇒ x = 0 or 18 m
$\therefore$ Range = 18 m

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