Question

The position coordinates of a projectile thrown from ground are given by y= 3t–5t² m & x = 4t m ( here t is in second, x is horizontal & y is vertical) horizontal range of the projectile is

Answer 1

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Answer 2

Answer:

The horizontal range of the projectile is $\dfrac{12}{5}\ m$

Explanation:

Given that,

The position of a particle is

Along x-axis

x=4t ....(I)

Along y-axis

y=3t-5t^2 .....(II)

From equation (I)

The value of t is

$t =\dfrac{x}{4}$

Now, put the value of t in equation (II)

$y=3\times\dfrac{x}{4}-5\times(\dfrac{x}{4})^2$

$y = \dfrac{3x}{4}-\dfrac{5x^2}{16}$.....(III)

For horizontal rang,

$y =0$

From equation (III)

$\dfrac{3x}{4}-\dfrac{5x^2}{16}=0$

$12x-5x^2=0$

$x = 0, \dfrac{12}{5}\ m$

We neglect the zero.

So,$x = \dfrac{12}{5}\ m$

Hence, The horizontal range of the projectile is $\dfrac{12}{5}\ m$