Question

23.
Two projectiles are shown in the figure IT, and T are the time periodu, and u are the velocities:
1)T2>T1
2)T1=T2
3)u1 >u2
4) u2=u1

Answer 1

From the figure we observe the following:

• Horizontal range, $\sf{R_1<R_2\quad\quad\dots(1)}$

• Angle of projection, $\sf{\theta_1>\theta_2\quad\quad\dots(2)}$

• Maximum height, $\sf{H_1=H_2\quad\quad\dots(3)}$

Consider (3).

$\sf{\longrightarrow H_1=H_2}$

$\sf{\longrightarrow \dfrac{(u_1)^2\sin^2\theta_1}{2g}=\dfrac{(u_2)^2\sin^2\theta_2}{2g}}$

Since initial speed is positive and angle of projection is acute,

$\sf{\longrightarrow u_1\sin\theta_1=u_2\sin\theta_2\quad\quad\dots(4)}$

$\sf{\longrightarrow\dfrac{u_1}{u_2}=\dfrac{\sin\theta_2}{\sin\theta_1}\quad\quad\dots(5)}$

But from (2),

$\sf{\longrightarrow\theta_1>\theta_2}$

Since both are acute and $\sf{\sin\theta\propto\theta\quad\!\forall\theta\in\left[0^o,\ 90^o\right],}$

$\sf{\longrightarrow\sin\theta_1>\sin\theta_2}$

$\sf{\longrightarrow\dfrac{\sin\theta_2}{\sin\theta_1}<1}$

Then (5) becomes,

$\sf{\longrightarrow\dfrac{u_1}{u_2}<1}$

$\sf{\longrightarrow u_1<u_2}$

Time of flight is given by,

$\sf{\longrightarrow T=\dfrac{2u\sin\theta}{g}}$

$\sf{\Longrightarrow T\propto u\sin\theta}$

So (4) can imply,

$\sf{\longrightarrow\underline{\underline{T_1=T_2}}}$

Hence (2) is the answer.