1.2 Features of projectile motion  (Page 7/7)

The time of flight is determined by considering vertical motion. It means that time of flight is dependent on speed and the angle of projection.

$$T=\frac{2u_y}{g}=\frac{2u\sin \theta }{g}$$

Maximum height is also determined, considering vertical motion. As such, maximum height also depends on the angle of projection.

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

Horizontal component, being component of velocity, depends on the angle of projection.

$$u_y=u\cos \theta$$

It is only the acceleration of projectile, which is equal to acceleration due to gravity and is, therefore, independent of the angle of projection. Hence, option (c) is correct.

The maximum heights, ranges and time of lights are compared, using respective formula as :

(i)Maximum Height

$$\Rightarrow \frac{H_1}{H_2}=\frac{u^2\sin {}^2{\theta }_1}{u^2\sin {}^2{\theta }_2}=\frac{\sin {}^2{30}^0}{\sin {}^2{60}^0}=\frac{{\left(\frac{1}{2}\right)}^2}{{\left(\frac{\sqrt{3}}{2}\right)}^2}=\frac{1}{3}$$

Thus, the maximum heights attained by two projectiles are unequal.

(ii) Range :

$$\Rightarrow \frac{R_1}{R_2}=\frac{u^2\sin 2{\theta }_1}{u^2\sin 2{\theta }_2}=\frac{{\sin }^2{60}^0}{{\sin }^2{120}^0}=\frac{{\left(\frac{\sqrt{3}}{2}\right)}^2}{{\left(\frac{\sqrt{3}}{2}\right)}^2}=\frac{1}{1}$$

Thus, the ranges of two projectiles are equal.

(iii) Time of flight

$$\Rightarrow \frac{T_1}{T_2}=\frac{2u\sin {\theta }_1}{2u\sin {\theta }_2}=\frac{\sin {30}^0}{\sin {60}^0}=\frac{\left(\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}}$$

Thus, the times of flight of two projectiles are unequal.

Hence, option (b) is correct.

We note that vertical component is negative, meaning that projectile is moving towards the ground. The vertical component of velocity 8 m above the ground is

$$v_y=-5m/s$$

The vertical displacement (y) from the maximum height to the point 8 m above the ground as shown in the figure can be obtained, using equation of motion.

$$v_y^2=u_y^2+2ay$$

Considering the point under consideration as origin and upward direction as positive direction.

$$\Rightarrow {\left(-5\right)}^2=0+2X-10Xh$$ $$\Rightarrow h=-25/20=-1.25m$$

Thus, the maximum height, H, attained by the projectile is :

$$\Rightarrow H=8+1.25=9.25m$$

Hence, option (c) is correct.

The projectile covers maximum range when angle of projection is equal to 45°. The maximum range "R" is given by :

$$\Rightarrow R=\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin {90}^0}{g}=\frac{u^2}{g}$$

On the other hand, the maximum height attained by the projectile for angle of projection, 45°, is :

$$\Rightarrow H=\frac{u_y^2}{2g}=\frac{u^2\sin {}^2{45}^0}{2g}=\frac{u^2}{4g}$$

Comparing expressions of range and maximum height, we have :

$$\Rightarrow R=4H$$

Hence, option (a) is correct.

The horizontal range of the projectile is given as :

$$R=\frac{u^2\sin 2\theta }{g}$$

In order to evaluate this expression, we need to know the angle of projection. Now, the initial part of the question says that the speed of a projectile at maximum height is half its speed of projection, "u". However, we know that speed of the projectile at the maximum height is equal to the horizontal component of projection velocity,

$$u\cos \theta =\frac{u}{2}$$ $$\Rightarrow \cos \theta =\frac{1}{2}=\cos {30}^0$$ $$\Rightarrow \theta ={30}^0$$

The required range is :

$$\Rightarrow R=\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin {60}^0}{g}=\frac{\sqrt{3}{u}^2}{2g}$$

Hence, option (b) is correct.

Here, we are required to find the product of times of flight. Let " $\theta$ " and " ${90}^0-\theta$ " be two angles of projections. The times of flight are given as :

$$T_1=\frac{2u\sin \theta }{g}$$

and

$$T_2=\frac{2u\sin \left({90}^0-\theta \right)}{g}=\frac{2u\cos \theta }{g}$$

Hence,

$$\Rightarrow T_1{T}_2=\frac{4u^2\sin \theta \cos \theta }{g^2}$$

But,

$$\Rightarrow R=\frac{2u^2\sin \theta \cos \theta }{g}$$

Combining two equations,

$$\Rightarrow T_1{T}_2=\frac{2R}{g}$$

Hence, option (c) is correct.

The magnitude of average velocity is given as :

$$v_{avg}=\frac{\text{Displacement}}{\text{Time}}$$

Now the displacement here is OA as shown in the figure. From right angle triangle OAC,

$$OA=\sqrt{\left(O{C}^2+B{C}^2\right)}=\sqrt{\{{\left(\frac{R}{2}\right)}^2+H^2\}}$$ $$\Rightarrow OA=\sqrt{\left(\frac{R^2}{4}+H^2\right)}$$

Hence, magnitude of average velocity is :

$$\Rightarrow v_{avg}=\frac{\sqrt{\left(\frac{R^2}{4}+H^2\right)}}{T}$$

Substituting expression for each of the terms, we have :

$$\Rightarrow v_{avg}=\frac{\sqrt{\left(\frac{u^4\sin {}^{2}2\theta }{4g^2}+\frac{u^4\sin {}^4\theta }{4g^2}\right)}}{\frac{2u\sin \theta }{g}}$$

vavg = u(3 u2cos + 1)/2 $$\Rightarrow v_{avg}=\frac{u\sqrt{\left(3u^2\cos \theta +1\right)}}{2}$$

Hence, option (d) is correct.

The angle of elevation of the highest point " $\theta$ " is shown in the figure. Clearly,

$$\tan \alpha =\frac{AC}{OC}=\frac{H}{\frac{R}{2}}=\frac{2H}{R}$$

Putting expressions of the maximum height and range of the flight, we have :

$$\Rightarrow \tan \alpha =\frac{2u^2{\sin }^2\theta Xg}{2gX{u}^2\sin 2\theta }=\frac{{\sin }^2\theta }{2\sin \theta \cos \theta }=\frac{\tan \theta }{2}$$

Hence, option (b) is correct.

The bullets can spread around in a circular area of radius equal to maximum horizontal range. The maximum horizontal range is given for angle of projection of 45°.

$$R_{\text{max}}=\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin 2X{45}^0}{g}=\frac{u^2\sin {90}^0}{g}=\frac{u^2}{g}$$

The circular area corresponding to the radius equal to maximum horizontal range is given as :

$$\Rightarrow A=\pi R_{\text{max}}^2=\pi X{\left(\frac{u^2}{g}\right)}^2=\frac{\pi u^4}{g^2}$$

Hence, option (c) is correct.