Question

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 1

$\checkmark$ 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

Answer 2

$\large{\underline{\underline{\sf{Given-}}}}$

• $\sf{y = 10x - \frac{5}{9} \: {x}^{2} }$

• $\sf{y=0}$

$\large{\underline{\underline{\sf{To \: Find-}}}}$

• $\sf{Range \: Of \: Projectile}$

$\large{\underline{\underline{\sf{Solution-}}}}$

$\implies \sf{0 = 10x - \frac{5}{9} \: {x}^{2} }$

$\implies \sf{10x = \frac{5}{9} \: {x}^{2} }$

$\implies \sf{x = \frac{10 \times 9}{5} }$

$\implies \sf{x = \frac{\cancel{90}}{\cancel{5}}}$

$\implies \sf{x = 18m}$

⛬ $\sf{Range \: Of \: Projectile \: is \: 18m}$