Projectile motion  (Page 6/6)

The horizontal component of acceleration is zero. As such, horizontal component of velocity is constant. On the other hand, vertical component of acceleration is “g”, which is a constant. Clearly, vertical component of velocity is not constant.

Hence, options (b) and (d) are correct.

Here, the vertical component of the velocity (3 m/s) is positive. Therefore, projectile is moving in positive y-direction. It means that the projectile is still ascending to reach the maximum height (the vertical component of velocity at maximum height is zero). It is though possible that the given velocity is initial velocity of projection. However, the same can not be concluded.

Hence, option (d) is correct.

The equation of projectile is given as :

$$y=x\tan \theta -\frac{g{x}^2}{u^2{\cos }^2\theta }$$

This is a quadratic equation in x. The correct choice is (b).

There is no component of acceleration in horizontal direction. The motion in this direction is a uniform motion and, therefore, covers equal horizontal distance in all parts of motion. It means that options (a) and (c) are incorrect.

In the vertical direction, projectile covers maximum distance when vertical component of velocity is greater. Now, projectile has greater vertical component of velocity near ground at the time of projection and at the time of return. As such, it covers maximum distance near the ground.

The correct choice is (d).

Average velocity is defined as the ratio of displacement and time. Since we treat projectile motion as two dimensional motion, we can find average velocity in two mutually perpendicualr directions and then find the resultant average velocity. Projectile motion, however, is a unique case of uniform acceleration and we can find components of average velcity by averaging initial and final values.

If $v_1$ and $v_2$ be the initial and final velocities, then the average velocity for linear motion under constant acceleration is defined as :

$$v_{avg}=\frac{v_1+v_2}{2}$$

Employing this relation in x-direction and making use of the fact that motion in horizontal direction is uniform motion, we have :

$$\Rightarrow v_{avgx}=\frac{u_x+v_x}{2}$$ $$\Rightarrow v_{avgx}=\frac{u\cos \theta +u\cos \theta }{2}=u\cos \theta$$

Similarly, applying the relation of average velocity in y-direction and making use of the fact that component of velocity in vertical direction reverses its direction on return, we have :

$$\Rightarrow v_{avgy}=\frac{u_y+v_y}{2}$$ $$\Rightarrow v_{avgy}=\frac{u\sin \theta -u\sin \theta }{2}=0$$

Hence, the resultant average velocity is :

$$\Rightarrow v_{avg}=v_{avgx}=u\cos \theta$$

Both initial and final speeds are equal. Hence, there is no change in speed during the motion of projectile.

Initial velocity is :

$$\mathbf{u}=u\cos \theta \mathbf{i}+u\sin \theta \mathbf{j}$$

Final velocity is :

$$\mathbf{v}=u\cos \theta \mathbf{i}-u\sin \theta \mathbf{j}$$

Change in velocity is given by :

$$\Rightarrow \text{Δ}\mathbf{v}=u\cos \theta \mathbf{i}-u\sin \theta \mathbf{j}-u\cos \theta \mathbf{i}-u\sin \theta \mathbf{j}$$ $$\Rightarrow \text{Δ}\mathbf{v}=-2u\sin \theta \mathbf{j}$$

where j is unit vector in y-direction.

We shall make use of the fact that horizontal component of projectile velocity does not change with time. Therefore, we can equate horizontal components of velocities at the time of projection and at the given instants and find out the speed at the later instant as required. Then,

$$u_x=v_x$$ $$\Rightarrow u\cos {60}^0=v\cos {30}^0$$ $$\Rightarrow \frac{10}{2}=\frac{\sqrt{3}v}{2}$$ $$\Rightarrow v=\frac{10}{\sqrt{3}}m/s$$

The displacement in x-direction is given as :

$$x=v_{x}t$$

Since $v_x$ is a constant, the "x-t" plot is a straight line passing through origin. Hence, option (a) is correct.

The displacement in y-direction is :

$$y=u_{y}t-\frac{1}{2}g{t}^2$$

This is an equation of parabola. The option (b), therefore, is incorrect.

The motion in horizontal direction is uniform motion. This means that component of velocity in x-direction is a constant. Therefore, " $v_x-t$ ” plot should be a straight line parallel to time axis. It does not pass through origin. Hence, option (c) is incorrect.

The motion in vertical direction has acceleration due to gravity in downward direction. The component of velocity in y-direction is :

$$v_y=u_y-gt$$

This is an equation of straight line having slope of "-g" and intercept " $u_y$ ". The " $v_y-t$ ” plot, therefore, is a straight line. Hence, option (d) is correct.