5.3 Features of projectile motion (application)  (Page 3/3)

Problem : The position of a projectile projected from the ground is :

$$\begin{array}{l}x=3t\\ y=(4t-2t^2)\end{array}$$

where “x” and “y” are in meters and “t” in seconds. The position of the projectile is (0,0) at the time of projection. Find the speed with which the projectile hits the ground.

Solution : When the projectile hits the ground, y = 0,

$$\begin{array}{l}0=(4t-2t^2)\\ \Rightarrow 2t^2-4t=t(2t-2)=0\\ \Rightarrow t=0,t=2s\end{array}$$

Here t = 0 corresponds to initial condition. Thus, projectile hits the ground in 2 s. Now velocities in two directions are obtained by differentiating given functions of the coordinates,

$$\begin{array}{l}{v}_x=\frac{đx}{đt}=3\\ v_y=\frac{đy}{đt}=4-4t\end{array}$$

Now, the velocities for t = 2 s,

$$\begin{array}{l}{v}_x=\frac{đx}{đt}=3m/s\\ v_y=\frac{đy}{đt}=4-4x2=-4m/s\end{array}$$

The resultant velocity of the projectile,

$$\begin{array}{l}v=\sqrt{({v_x}^2+{v_y}^2)}=\sqrt{\{3^2+{(-4)}^2\}}=5m/s\end{array}$$

Problem : A projectile, thrown from the foot of a triangle, lands at the edge of its base on the other side of the triangle. The projectile just grazes the vertex as shown in the figure. Prove that :

$$\tan \alpha +\tan \beta =\tan \theta$$

where “θ” is the angle of projection as measured from the horizontal.

Solution : In order to expand trigonometric ratio on the left side, we drop a perpendicular from the vertex of the triangle “A” to the base line OB to meet at a point C. Let x,y be the coordinate of vertex “A”, then,

$$\tan \alpha =\frac{AC}{OC}=\frac{y}{x}$$

and

$$\tan \beta =\frac{AC}{BC}=\frac{y}{\left(R-x\right)}$$

Thus,

$$\tan \alpha +\tan \beta =\frac{y}{x}+\frac{y}{\left(R-x\right)}=\frac{yR}{x\left(R-x\right)}$$

Intuitively, we know the expression is similar to the expression involved in the equation of projectile motion that contains range of projectile,

$$\Rightarrow y=x\tan \theta \left(1-\frac{x}{R}\right)$$

$$\Rightarrow \tan \theta =\frac{yR}{x\left(R-x\right)}$$

Comparing equations,

$$\Rightarrow \tan \alpha +\tan \beta =\tan \theta$$

Problem : A projectile is projected at angle “θ” from the horizontal at the speed “u”. If an acceleration of “g/2” is applied to the projectile due to wind in horizontal direction, then find the new time of flight, maximum height and horizontal range.

Solution : The acceleration due to wind affects only the motion in horizontal direction. It would, therefore, not affect attributes like time of flight or maximum height that results exclusively from the consideration of motion in vertical direction. The generic expressions of time of flight, maximum height and horizontal range of flight with acceleration are given as under :

$$T=\frac{2u_y}{g}$$

$$H=\frac{u_y^2}{2g}=\frac{g{T}^2}{4}$$

$$R=\frac{u_x{u}_y}{g}$$

The expressions above revalidate the assumption made in the beginning. We can see that it is only the horizontal range that depends on the component of motion in horizontal direction. Now, considering accelerated motion in horizontal direction, we have :

$$x=R\prime =u_{x}T+\frac{1}{2}{a}_x{T}^2$$

$$\Rightarrow R\prime =u_{x}T+\frac{1}{2}\left(\frac{g}{2}\right)T^2$$

$$R\prime =R+H$$

Problem : A projectile is projected at angle “θ” from the horizontal at the speed “u”. If an acceleration of g/2 is applied to the projectile in horizontal direction and a deceleration of g/2 in vertical direction, then find the new time of flight, maximum height and horizontal range.

Solution : The acceleration due to wind affects only the motion in horizontal direction. It would, therefore, not affect attributes resulting exclusively from the consideration in vertical direction. It is only the horizontal range that will be affected due to acceleration in horizontal direction. On the other hand, deceleration in vertical direction will affect all three attributes.

1: Time of flight

Let us work out the effect on each of the attribute. Considering motion in vertical direction, we have :

$$y=u_{y}T+\frac{1}{2}{a}_y{T}^2$$

For the complete flight, y = 0 and t = T. Also,

$$a_y=-\left(g+\frac{g}{2}\right)=-\frac{3g}{2}$$

Putting in the equation,

$$\Rightarrow 0=u_{y}T-\frac{1}{2}X\frac{3g}{2}X{T}^2$$

Neglecting T = 0,

$$\Rightarrow T=\frac{4u_y}{3g}=\frac{4u\sin \theta }{3g}$$

2: Maximum height

For maximum height, $v_y=0$ ,

$$0=u_y^2-2X\frac{3g}{2}XH$$

$$H=\frac{u_y^2}{3g}=\frac{u^2{\sin }^2\theta }{3g}$$

2: Horizontal range

Now, considering accelerated motion in horizontal direction, we have :

$$x=R=u_{x}T+\frac{1}{2}{a}_x{T}^2$$

$$R=u_x\left(\frac{4u_y}{g}\right)+\frac{1}{2}\left(\frac{g}{2}\right){\left(\frac{4u_y}{g}\right)}^2$$

$$\Rightarrow R=\left(\frac{4u_y}{g}\right)[u_x+\frac{1}{2}\left(\frac{g}{2}\right)\left(\frac{4u_y}{g}\right)]$$

$$\Rightarrow R=\left(\frac{4u_y}{g}\right)\{u_x+u_y\}$$

$$\Rightarrow R=\left(\frac{4u^2\sin \theta }{g}\right)[\cos \theta +\sin \theta ]$$