Question

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Answer 1

• The platform is pulled with a horizontal acceleration $\sf{a.}$ So the particle projected will experience a pseudo acceleration $\sf{a}$ opposite to the direction of the pulling.
• If the particle comes back to the starting point after projection, the horizontal velocity of the particle should be opposite to horizontal acceleration.
• Let the particle projected with initial speed $\sf{u}$ at an angle $\theta$ with the horizontal.

So the initial velocity can be,

• $\vec{\sf{u}}=\sf{u\cos\theta\ \hat i+u\sin\theta\ \hat j}$

The net acceleration on the particle will be,

• $\vec{\sf{a}}=\sf{-a\ \hat i-g\ \hat j}$
• where $\sf{\hat i}$ represents the unit vector along the direction of pulling of the platform and $\sf{\hat j}$ is the unit vector along upward direction.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\circle*{2}}\put(0,0){\vector(-1,1){20}}\put(0,0){\vector(-1,0){20}}\put(0,0){\vector(0,1){20}}\put(0,0){\vector(1,0){10}}\put(0,0){\vector(0,-1){15}}\put(-23.5,21){\sf{u}}\put(-25,-4){ \sf{u\cos\theta} }\put(1.5,20){ \sf{u\sin\theta} }\qbezier(-4,0)(-4,2)(-2.5,2.6)\put(-7.5,1){ \theta }\put(50,0){\vector(-1,0){10}}\put(50,0){\vector(0,1){10}}\put(11.5,-0.5){\sf{a}}\put(-1,-18){\sf{g}}\put(40,-4){\sf{x}}\put(52,9){\sf{y}}\end{picture}$

• As the particle comes back to the starting point after the projection, the net displacement is zero vector.

So by second equation of motion:-

$\longrightarrow \vec{\sf{0}}=\sf{\left(u\cos\theta\ \hat i+u\sin\theta\ \hat j\right)t+\dfrac{1}{2}\left(-a\ \hat i-g\ \hat j\right)t^2}$

$\longrightarrow \sf{0\ \hat i+0\ \hat j}=\sf{\left(u\cos\theta\,t-\dfrac{1}{2}at^2\right)\,\hat i+\left(u\sin\theta\,t-\dfrac{1}{2}gt^2\right)\,\hat j}$

Equating corresponding components,

$\sf{\longrightarrow u\cos\theta-\dfrac{1}{2}at^2=0}$

$\sf{\longrightarrow u\cos\theta=\dfrac{1}{2}at^2}\quad\quad\dots(1)}$

and,

$\sf{\longrightarrow u\sin\theta-\dfrac{1}{2}gt^2=0}$

$\sf{\longrightarrow u\sin\theta=\dfrac{1}{2}gt^2}\quad\quad\dots(2)}$

Dividing (2) by (1),

$\sf{\longrightarrow \dfrac{u\sin\theta}{u\cos\theta}=\dfrac{\left(\dfrac{1}{2}gt^2\right)}{\left(\dfrac{1}{2}at^2\right)}}$

$\sf{\longrightarrow \tan\theta=\dfrac{g}{a}}$

Taking $\sf{g=10\ m\,s^{-2}}$ and $\sf{a=5\ m\,s^{-2},}$

$\sf{\longrightarrow\underline{\underline{\tan\theta=2}}}$

Answer 2

$\large\underline\color{skyblue}\sf Correct \: Question:-$

A platform is pulled with a constant acceleration a; A particle is projected from the platform at angle θ with the horizontal with respect to the platform as shown in the figure. The value of tanθ is such that particle again come to the starting point on the platform is

$\small\tt(a = 5m/{s}^{2} ) \: use \: g \: = 10m/ {s}^{2}$

$\large\underline\color{skyblue}\sf Answer :-$

Here we need to make two equations one in x direction and another one in y direction.

The motion of particle in x direction:

$\tt \: S_x=0$

From the equations for kinematics:

$\bf \: S_x = u + \frac{1}{2} {at}^{2}$

$\tt \blue {Here,}$

$\begin{gathered} \sf \: S_x \: = distance \: in \: the \: x \: direction \\ \sf u \: = initial \: velocity \\ \sf a \: = acceleration \\ \sf \: t \: = time \end{gathered}$

$\bf\implies \: S_x = u \: cos \theta \: - \frac{1}{2} gt ^{2}$

Write the above equation in terms of t

$\sf \implies \: u \: cos \theta \: = \frac{1}{2} {gt}^{2} \\ \sf \implies \frac{2 \: u \: cos \theta}{g} = {t}^{2}$

Similarly, in the y direction,

$\tt \:Sy = 0$

$\sf \: Sy = u \: + \frac{1}{2} \: {at}^{2}$

$\bf \: \implies \: Sy = u \: sin \theta \: - \frac{1}{2} {gt}^{2}$

Write the above equation in terms of t

$\sf\implies \: u \: sin \theta \: = \frac{1}{2} \: {gt}^{2} \\ \sf \implies \frac{2u \: sin \theta}{g} = {t}^{2}$

$\sf \: Now \: Divide \: \frac{2u \: sin \theta}{9} = {t}^{2} \: \: by \: \: \frac{2u \: cos \theta}{g} = {t}^{2}$

$\sf \large\implies \: \frac{2u \: sin \theta}{g} \times \frac{9}{2u \: cos \theta} = \frac{ {t}^{2} }{ {t}^{2} }$

Cancel out the common factors

$\sf \large\implies \: \frac{sin \theta}{1} \times \frac{1}{cos \theta} = 1$

Solve the above equation

$\sf\implies \: \frac{sin \theta}{cos \theta} = 1 \\ \\ \sf \implies \: tan \theta = 1$

We know that tanθ is equal to perpendicular upon base :

$\sf \ \implies \: tan \theta = \large\frac{g}{a}$

put in the given value in the above equation

$\sf\implies \: tan \theta = \frac{10}{5} \\ \sf\implies tan \theta = 2$

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