Question

define projectile.find the relation for time of flight, maximum height,horizontal range and maximum horizontal.range when projectile fired at some angle with horizontal. ​

Answer 1

Answer:

A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration.

Explanation:

Answer 2

Answer:

$T( time of flight )=\frac{(2 u \sin \theta)}{g} \\R( Horizontal range )=\frac{u^{2}(2 \sin \theta \cos \theta)}{g} \\H( Maximum height )=\frac{u^{2} \sin ^{2} \theta}{2 g}$

Explanation:

Projectile motion is a form of Motion experienced by an object or particle that is thrown near the Earth's surface and moves along a curved path under the action of gravity only .

The graph for projectile motion :

Terms related to projectile motion :

• The path of a projectile is called Trajectory.
• The Horizontal Range of a Projectile is defined as the horizontal displacement of a projectile when the displacement of the projectile in the y-direction is zero.
• Vertical range is the maximum vertical distance a projectile can reach. It is the same as the maximum vertical displacement denoted by (H). It is also known as Maximum height .

Let a projectile move with a velocity $u$ which is inclined with the horizontal at angle of $\theta$. The velocity after time ' $t$ ' will be

$\begin{aligned}&v=v_{x} \hat{i}+v_{y} \hat{j} \\&v=u \cos \theta \hat{i}+(u \sin \theta-g t) \hat{j} \\&\left(\text { Since }: v_{y}=u \sin \theta-g t\right)\end{aligned}$

At maximum height the projectile will only have horizontal component that is

$\begin{aligned}&\mathrm{v}_{\mathrm{x}}=\mathrm{u} \cos \theta \\&\mathrm{v}_{\mathrm{y}}^{2}-\mathrm{u}_{\mathrm{y}}^{2}=2 \mathrm{ay} \\&\left.\mathrm{v}_{\mathrm{y}}=0 \text { (at max height } \mathrm{H}\right) \\&\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta \\&\text { ay }=-\mathrm{gH} \\&\mathrm{P} \text { utting these values, } \\&0=(\mathrm{usin} \theta)^{2}-2 \mathrm{gH} \\&\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\end{aligned}$

First, there are some essential equation regarding one-dimension motion with constant acceleration:

$\begin{aligned}&v=v_{0}+a t \\&x=x_{0}+v_{0} t+\frac{1}{2} a t^{2} \\&v^{2}-v_{0}^{2}=2 a s\end{aligned}$

(where $v$ is the final velocity, $v_{0}$ is the initial velocity, $\mathrm{x}$ is the displacement, $s$ is the distance the object travels along the direction of acceleration)

(All this set of equations can be proved with simple calculus)

Now apply to two-dimension motion, Set up the projectile problem as shown below,(1) to find velocity components:

On the x-axis: $u_{x}=u_{0} \cos (\theta)$ (since there is no acceleration on $x$-axis)

On the y_axis: $u_{y}=u_{0} \sin (\theta)-g t (sine \left.a_{y}=-g\right)$

Apply (3) on the y axis, note that when the object reaches maximum height, $v_{y}=0$

$\Rightarrow 0^{2}-\left(u_{0} \sin (\theta)\right)^{2}=2(-g) H \\ \Rightarrow H=\frac{\left(u_{0} \sin (\theta)\right)^{2}}{2 g}$

Now, calculate the total flight time of the object, at $t=0 u_{y}=u_{0} \sin (\theta)$, when the object touches the ground, $u_{y}=-u_{0} \sin (\theta)$

Apply (1): $-u_{0} \sin (\theta)=u_{0} \sin (\theta)-g t_{1}$

$\rightarrow t_{1}=\frac{2 u_{0} \sin (\theta)}{g}$

To find the horizontal range:

$R=u_{x} \cdot t_{1}=u_{0} \cos (\theta) \frac{2 u_{0} \sin (\theta)}{a}$

which finally gives, $R=\frac{u_{0}^{2}}{g} 2 \sin (\theta) \cos (\theta)$

So the relation between $\mathrm{H} , \mathrm{R}$is

$R=4 H \frac{\cos (\theta)}{\sin (\theta)}$