Question

A particle is projected in vertical plane follows the path 3y=4x-15x square horizontal range is?

Answer 1

Equation of trajectory of an projectile which is projected at some angle with horizontal is given by

$y = xtan\theta(1 - \frac{x}{R})$

here

$\theta$ = angle with horizontal

R = horizontal range

now we have equation given as

$3y = 4x - 15x^2$

now we have

$3y = 4x(1 - \frac{15x}{4})$

$y = \frac{4}{3}x (1 - \frac{x}{0.267})$

now if we compare it with the equation of trajectory then we will have

$tan\theta = \frac{4}{3}$

R = 0.267 meter

so range of the projectile will be 0.267 m

Answer 2

Answer:4/15

Explanation:Since its horizintal range

y=0

4x=15x^2

x=4/15m

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