Question

A platform is pulled with horizontal acceleration a. A particle is
projected from the platform at an angle with the horizontal
with respect to the platform as shown in figure. The value of
tano such that particle again come to the starting point on the
platform is
(a = 5 m s 2 and g = 10 m s-?)
0
a>

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 $\sf{a=5\ m\,s^{-2},}$

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