the equation of tragectory of particle is y= 10x-5/9 x^2 where x and y

1.	the equation of tragectory of particle is y= 10x-5/9 x^2 where x and y are in mts, find the range of projectile
Answer» $\checkmark$ Given: the equation of TRAJECTORY of particle is $\sf \longrightarrow y=10x-\dfrac{5}{9}x^{2}$ where x and y are in METRES $\checkmark$ To Find: The range of projectile $\checkmark$ SOLUTION: We know that, Horizontal Range of projectile is displacement covered in X axis HENCE, For Range, y = 0 According to the question, We are asked to find the range of projectile So, We must find "x" Here, y = 0 Hence, SUBSTITUTING the value, We get, $\blue{\sf \implies y=10x-\dfrac{5}{9}x^{2}}$ $\green{\sf \implies 0=10x-\dfrac{5}{9}x^{2}}$ $\orange{\sf \implies 10x=\dfrac{5}{9}x^{2}}$ On further simplification, We get, $\red{\sf \implies x=\dfrac{10 \times 9}{5} \ m}$ Hence, $\pink{\sf \implies x= 18 \ m}$ Therefore, $\star$ Range of projectile = 18 m