A projectile is fired at an angle with the horizontal. 1. Show that its trajectory is a parabola. 2. Obtain expression for: 1. The maximum height attained. 2. The time of its flight and 3. The horizontal range. 3. At what value of is the horizontal range maximum? 4. Prove that, for a given velocity of projection, the horizontal range is same for and (90^ - ).

When a body is projected in the air in any direction, then the body is called a projectile. 1. Suppose a body is projected with velocity u at an angle with the horizontal, P(x, y) is any point on its trajectory at time t. Horizontal component of velocity is unaffected by gravity, but the vertical component (u ) changes due to gravity x=(u )t. y=(u )t-1 2 gt^2 =u x u -1 2 g ( x u )^2 y=x - gx^2 2u^2 ^2 (i) It represents the equation of a parabola, hence the path followed by a projectile is a parabola. 2. The greatest vertical distance attained by the projectile above the horizontal plane from the point of projection is called maximum height. Maximum height, LN = H 1. At maximum height, v = 0 v^2-u_y^2=-2gH, where u_y=u or (u )^2=2gH or H= u^2 ^2 2g 2. At maximum height, v = 0 0=u -gt or t= u g But time of flight, T=2t= 2u g 3. When the body returns to the same horizontal level y = 0 0=x - gx^2 2u^2 ^2 [From (i)] or x = gx^2 2u^2 ^2 or x= 2u^2 g = u^2 2 g But coordinates of M are (R, 0). Putting x = R, we have, R= u^2 2 g . 3. =45^ 4. When an object is projected with velocity u making an angle with horizontal direction. R_1= u^2 2 g (i) When an object is projected with u making an angle (90^ - ) R_2= u^2 2(90^ - ) g = u^2 g (180^ -2 ) = u^2 g 2 (ii) form (i) and (ii) R_1 = R_2 The horizontal range is same for two complementary angles.

A projectile is fired at an angle with the horizontal. 1. Show th…

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Question

A projectile is fired at an angle with the horizontal.

Show that its trajectory is a parabola.
Obtain expression for:

The maximum height attained.
The time of its flight and
The horizontal range.

At what value of $\theta$ is the horizontal range maximum?
Prove that, for a given velocity of projection, the horizontal range is same for $\theta$ and $(90^\circ-\theta).$

✓

Answer

When a body is projected in the air in any direction, then the body is called a projectile.

Suppose a body is projected with velocity u at an angle $\theta$ with the horizontal, P(x, y) is any point on its trajectory at time t.

Horizontal component of velocity is unaffected by gravity, but the vertical component $(\text{u}\sin\theta)$ changes due to gravity
$\therefore\ \text{x}=(\text{u}\cos\theta)\text{t}.$
$\text{y}=(\text{u}\sin\theta)\text{t}-\frac{1}{2}\text{gt}^2$
$=\text{u}\sin\theta\times\frac{\text{x}}{\text{u}\cos\theta}-\frac{1}{2}\text{g}\Big(\frac{\text{x}}{\text{u}\cos\theta}\Big)^2$
$\text{y}=\text{x}\tan\theta-\frac{\text{gx}^2}{2\text{u}^2\cos^2\theta}\dots(\text{i})$
It represents the equation of a parabola, hence the path followed by a projectile is a parabola.

The greatest vertical distance attained by the projectile above the horizontal plane from the point of projection is called maximum height.

Maximum height, LN = H

At maximum height,

v = 0
$\therefore\ \text{v}^2-\text{u}_\text{y}^2=-2\text{gH},$ where
$\text{u}_\text{y}=\text{u}\sin\theta$
or $(\text{u}\sin\theta)^2=2\text{gH}$
or $\text{H}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$

At maximum height,

v = 0
$\therefore\ 0=\text{u}\sin\theta-\text{gt}$
or $\text{t}=\frac{\text{u}\sin\theta}{\text{g}}$
But time of flight,
$\text{T}=2\text{t}=\frac{2\text{u}\sin\theta}{\text{g}}$

When the body returns to the same horizontal level y = 0

$\therefore\ 0=\text{x}\tan\theta-\frac{\text{gx}^2}{2\text{u}^2\cos^2\theta}$ [From (i)]
or $\text{x}\tan\theta=\frac{\text{gx}^2}{2\text{u}^2\cos^2\theta}$
or $\text{x}=\frac{2\text{u}^2\sin\theta\cos\theta}{\text{g}}=\frac{\text{u}^2\sin2\theta}{\text{g}}$
But coordinates of M are (R, 0). Putting x = R,
we have,
$\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}.$

$\theta=45^\circ$
When an object is projected with velocity u making an angle $\theta$ with horizontal direction.

$\text{R}_1=\frac{\text{u}^2\sin2\theta}{\text{g}}\dots(\text{i})$
When an object is projected with u making an angle $(90^\circ-\theta)$
$\text{R}_2=\frac{\text{u}^2\sin2(90^\circ-\theta)}{\text{g}}$
$=\frac{\text{u}^2}{\text{g}}\sin(180^\circ-2\theta)$
$=\frac{\text{u}^2}{\text{g}}\sin2\theta\dots(\text{ii})$
form (i) and (ii) $R_1 = R_2$
$\therefore$ The horizontal range is same for two complementary angles.

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