Question

the equation of motion of a projectile is y=12x - 5/9x^2, where x and y are in meters. the range of the projectile is :
a) 36m
b) 30.6m
c)21.6m
d) 12.4m

Answer 1

Answer:

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Answer 2

Given:

Equation of motion of projectile,

$\bold{y=12x-\frac{5}{9}x^{2} }$, where x and y are in meters.

To Find:

Range of projectile.

Solution:

When we project a object i.e. in the initial stage height of the object is 0, and at the final state the object will be in ground, then also its height is zero. That means in both the cases $\bold{y=0}$.

Now, to find the range (maximum value of x) we have to take $\bold{y=0}$

∴The given equation becomes,

$\bold{12x-\frac{5}{9}x^{2} =0}$ ….. (1)

Now, find the value of x from equation (1)

∴$\bold{x(12-\frac{5}{9}x) =0}$

$\bold{=>x=0m\hspace{1cm} or\hspace{1cm} x=\frac{12\times 9}{5} = 21.6m }$

Here, $\bold{x=0}$ is for initial stage

And at final point where x is maximum (range of projectile) its value becomes $\bold{21.6m}$.

∴Range of the projectile is 21.6 m,

∴ Option (c) is the correct option