Ncert Exercises & Solutions

Two particles are projected in the air with speed $v_0$ , at angles $\theta_1$ and $\theta_2$ to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then:

(a)	the angle of the projection: $\theta_1>\theta_2$
(b)	the time of flight: $T_1>T_2$
(c)	the horizontal range: $R_1>R_2$
(d)	the total energy: $U_1>U_2$

Choose the correct option:
1. (a), (c), (d)
2. (a), (c)
3. (b), (c), (d)
4. (a), (b)

(4) Hint: The maximum height depends on the angle of projection.

_Step_{1: Find a relation between angles.z}

_{We know that maximum height reached by a projectile,}

_{$\begin{matrix} H = \frac{u^{2} {sin}^{2} θ}{2 g} H_{1} = \frac{v_{0}^{2} {sin}^{2} θ_{1}}{2 g} (for first particle) H_{2} = \frac{v_{0}^{2} {sin}^{2} θ_{2}}{2 g} (for second particle) \end{matrix}$ $\begin{array}{l} H = \frac{u^{2} \sin^{2} θ}{2 g} \\ H_{1} = \frac{v_{0}^{2} \sin^{2} θ_{1}}{2 g} (for first particle) \\ H_{2} = \frac{v_{0}^{2} \sin^{2} θ_{2}}{2 g} (for second particle) \end{array}$}

_{According to the question, we know that}

_{$\begin{matrix} \Rightarrow \frac{H_{1} > H_{2}}{2 g} > \frac{v_{0}^{2} {sin}^{2} θ_{1}}{2 g} > \frac{v_{0}^{2} {sin}^{2} θ_{2}}{2 g} \Rightarrow {sin}^{2} θ_{1} > {sin}^{2} θ_{2} \end{matrix}$ $\begin{array}{l} \Rightarrow \frac{H_{1} > H_{2}}{2 g} > \frac{v_{0}^{2} \sin^{2} θ_{1}}{2 g} > \frac{v_{0}^{2} \sin^{2} θ_{2}}{2 g} \\ \Rightarrow \sin^{2} θ_{1} > \sin^{2} θ_{2} \end{array}$

$\begin{matrix} \Rightarrow & {sin}^{2} θ_{1} - {sin}^{2} θ_{2} > 0 \Rightarrow & (sin θ_{1} - sin θ_{2}) (sin θ_{1} + sin θ_{2}) > 0 \end{matrix}$ $\begin{array}{lr} \Rightarrow & \sin^{2} θ_{1} - \sin^{2} θ_{2} > 0 \\ \Rightarrow & (\sin θ_{1} - \sin θ_{2}) (\sin θ_{1} + \sin θ_{2}) > 0 \end{array}$

$\begin{matrix} Thus, either & sin θ_{1} + sin θ_{2} & > 0 \Rightarrow & sin θ_{1} - sin θ_{2} & > 0 \Rightarrow & sin θ_{1} & > sin θ_{2} or θ_{1} > θ_{2} Time of fight, & T & = \frac{2 usin θ}{g} = \frac{2 v_{0} sin θ}{g} Thus, & T_{1} = \frac{2 v_{0} sin θ_{1}}{g} T_{2} & = \frac{2 v_{0} sin θ_{2}}{g} \end{matrix}$ $\begin{array}{lrl} Thus, either & \sin θ_{1} + \sin θ_{2} & > 0 \\ \Rightarrow & \sin θ_{1} - \sin θ_{2} & > 0 \\ \Rightarrow & \sin θ_{1} & > \sin θ_{2} or θ_{1} > θ_{2} \\ Time of fight, & T & = \frac{2 usin θ}{g} = \frac{2 v_{0} \sin θ}{g} \\ Thus, & T_{1} = \frac{2 v_{0} \sin θ_{1}}{g} \\ T_{2} & = \frac{2 v_{0} \sin θ_{2}}{g} \end{array}$}

_{(Here. $T_{1}$ $T_{1}$ = Time of flight of first particle and $T_{2}$ $T_{2}$ = Time of flight of second particle).}

Step 2: Find relation between time of flight and the range.

$\begin{matrix} As, & sin θ_{1} > sin θ_{2} Hence, & T_{1} > T_{2} \end{matrix}$ $\begin{array}{lc} As, & \sin θ_{1} > \sin θ_{2} \\ Hence, & T_{1} > T_{2} \end{array}$

We know that,

$\begin{matrix} Range, R = \frac{u^{2} sin 2 θ}{g} = \frac{v_{0}^{2} sin 2 θ}{g} \begin{matrix} A_{1} = Range of first particle = \frac{u_{0}^{2} sin 2 θ_{1}}{g} R_{2} = Range of second particle = \frac{v_{0}^{2} sin 2 θ_{2}}{g} \end{matrix} \end{matrix}$ $\begin{array}{l} Range, R = \frac{u^{2} \sin 2 θ}{g} = \frac{v_{0}^{2} \sin 2 θ}{g} \\ \begin{matrix} A_{1} = Range of first particle = \frac{u_{0}^{2} \sin 2 θ_{1}}{g} \\ R_{2} = Range of second particle = \frac{v_{0}^{2} \sin 2 θ_{2}}{g} \end{matrix} \end{array}$

Given,

$\begin{matrix} \cdot sin θ_{1} > sin θ_{2} \Rightarrow & sin 2 θ_{1} > sin 2 θ_{2} \Rightarrow & \frac{R_{1}}{R_{2}} = \frac{sin 2 θ_{1}}{sin 2 θ_{2}} > 1 \Rightarrow & R_{1} > R_{2} \end{matrix}$ $\begin{array}{lc} \cdot \sin θ_{1} > \sin θ_{2} \\ \Rightarrow & \sin 2 θ_{1} > \sin 2 θ_{2} \\ \Rightarrow & \frac{R_{1}}{R_{2}} = \frac{\sin 2 θ_{1}}{\sin 2 θ_{2}} > 1 \\ \Rightarrow & R_{1} > R_{2} \end{array}$

Step 3: Find a relation between the total energies.

Total energy for the first particle.

$\begin{matrix} U_{1} = KE + PE = \frac{1}{2} m_{1} v_{0}^{2} (This value will be constant throughout the journey) U_{2} = KE + PE = \frac{1}{2} m_{2} v_{0}^{2} (Total energy for the second particle) \end{matrix}$ $\begin{array}{l} U_{1} = KE + PE = \frac{1}{2} m_{1} v_{0}^{2} \\ (This value will be constant throughout the journey) \\ U_{2} = KE + PE = \frac{1}{2} m_{2} v_{0}^{2} (Total energy for the second particle) \end{array}$

Total energy for the particle

$\begin{matrix} m_{1} = m_{2} then U_{1} = U_{2} m_{1} > m_{2} then U_{1} > U_{2} m_{1} < m_{2}, then U_{1} < U_{2} \end{matrix}$ $\begin{array}{l} m_{1} = m_{2} then U_{1} = U_{2} \\ m_{1} > m_{2} then U_{1} > U_{2} \\ m_{1} < m_{2}, then U_{1} < U_{2} \end{array}$

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