Question

if y and x are representing vertical and horizontal displacements from reference point kept in horizontal and trajectory of projectile is y= 2x - 6x². then the maximum height and range of projectile will be​

Answer 1

Explanation:

Let the velocity of projection of the projectile in x-y plane from the origin (0,0) be u with angle of projection

$α$

with the horizontal direction (x-axis).

The vertical component of the velocity (along y-axis) of projection is

$u</p><p>sin</p><p>α$

and the horizontal component is

$u</p><p>cos</p><p>α$

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical displacement h will be zero. So applying the equation of motion under gravity we can write

$h</p><p>=</p><p>u</p><p>sin</p><p>α</p><p>×</p><p>T</p><p>+</p><p> \frac{1}{2} </p><p>g</p><p>T</p><p>2$

$⇒</p><p>0</p><p>=</p><p>u</p><p>×</p><p>T</p><p>−</p><p>1</p><p>2</p><p>×</p><p>g</p><p>×</p><p>T</p><p>2$

$where \: g= \: acceleration \: due \: to \: gravity$

since,

$\frac{</p><p>T</p><p>=</p><p>2</p><p>u</p><p>sin</p><p>α</p><p>}{g}$

The horizontal displacement during this T sec or the Range ,

$R</p><p>=</p><p>u</p><p>cos</p><p>α</p><p>×</p><p>T</p><p>$

$⇒</p><p>R</p><p>=</p><p>2</p><p>u</p><p>2</p><p>sin</p><p>α</p><p>cos</p><p>α</p><p>g$

...

$(</p><p>1</p><p>)</p><p>$

Let the position of the projectile in x-y plane after t sec of its projection be

$(</p><p>x</p><p>,</p><p>y</p><p>)$

The horizontal displacement during t sec

$x</p><p>=</p><p>u</p><p>cos</p><p>α</p><p>×</p><p>t</p><p> \: \: \: \: \: ...</p><p>.</p><p>(</p><p>2</p><p>)$

And the vertical displacement during t sec

$y</p><p>=</p><p>u</p><p>sin</p><p>α</p><p>×</p><p>t</p><p>−</p><p>1</p><p>2</p><p>×</p><p>g</p><p>×t2</p><p> \: \: \: \: \: \: \: \: \: \: \: ...</p><p>.</p><p>.</p><p>(</p><p>3</p><p>)</p><p>$

Combining (1) and (2) we get

$y</p><p>=</p><p>u</p><p>sin</p><p>α</p><p>×</p><p> \frac{ x</p><p> }{ u</p><p>cos</p><p>α}</p><p>−</p><p> \frac{g</p><p>x</p><p>2}{2</p><p>u</p><p>2</p><p>cos</p><p>2</p><p>α} </p><p></p><p>$

⇒

y

=

x

tan

α

−

x

2

2

u

2

cos

2

α

g

...

.

(

4

)

This is the equation of the trajectory,we obtained.

Now in the given problem the equation of the trajectory is

y

=

2

x

(

1

−

x

40

)

⇒

y

=

2

x

−

x

2

20

...

...

(

5

)

Comparing (4) and (5) we get

tan

α

=

2

...

.

.

(

6

)

and

2

u

2

cos

2

α

g

=

20

...

.

.

(

7

)

Multiplying (6) by (7) we get

tan

α

×

2

u

2

cos

2

α

g

=

2

×

20

⇒

2u/2sinαcosαg=40

⇒

R = 40

So Range of the projectile

R

=

40

m