Question

A particle projected from origin under uniform gravity equation of its path is given as y is equal to x minus b x square find its range

Answer 1

Answer:

Given:

The equation of a projectile is as follows

y = x - bx²

To find:

Range of the projectile

Definitions:

Range is referred to the maximum Horizontal distance covered by a projectile object when thrown at a certain velocity and at a particular angle.

Concept:

The projectile object starts from the ground and ends on the ground as soon as it covers the range.

So we can say that y-axis displacement is zero. If we put the value of y = 0 in that equation, we will get the corresponding value of x , which will denote Range.

Calculation:

y = x - bx²

Putting y = 0

=> 0 = x - bx²

=> x (1 - bx ) = 0

So, 1 - bx = 0

=> bx = 1

=> x = (1/b).

So the final answer is :

$\huge{\boxed{range = \frac{1}{b}}}$

Answer 2

$\Huge{\underline{\underline{\red{\sf{Answer :}}}}}$

As the Path of a Projectile is given,

$\Large \longrightarrow {\underline{\boxed{\sf{y \: = \: x \: - \: bx^2}}}}$

As we are given object is started from rest and it again goes in state of rest after reaching ground. So, y = 0

As, Y-axis is 0.

And x be The Horizontal Range.

Now, Put Values

$\large \rightarrow {\sf{0 \: = \: x \: - \: bx^2}}$

$\large \rightarrow {\sf{0 \: = \: x(1 \: - \: bx) }}$

$\large \rightarrow {\sf{1 \: - \: bx \: = \: 0}}$

$\large \rightarrow {\sf{1 \: = \: bx}}$

$\LARGE \longrightarrow {\boxed{\sf{x \: = \: \frac{1}{b}}}}$

∴ Range is 1/b

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Some Important Formulas :

[For Horizontal Projection]

• Path of Projectile , y = kx²

• Time of Flight, T = √2h/g

• Horizontal Range, R = u√2h/g

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