Question

A man is travelling on a flatcar which is moving up a plane that is inclined at angle $\theta$ to the horizontal with a speed of $5\text{m/s}$ $(\cos \theta =\frac{4}{5})$. He throws a ball towards a stationary hoop located above the incline in such a way that the ball moves parallel to the slope of the incline while going through the centre of the hoop. The centre of the hoop is $4\text{m}$ high from the man's hand, perpendicular to the incline. The time taken (in seconds) by the ball to reach the hoop is [Take $g=10{\text{m/s}}^2$]

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Answer 1

Considering the motion of ball separately along the directions $x$ and $y$ as given in question.


Here, $\alpha$ is the angle made by the ball with the inclined plane and $\theta$ is the angle made by the inclined plane with the horizontal
$a_x=-g\sin \theta ,a_y=-g\cos \theta$
$u_x=u\cos \alpha ,u_y=u\sin \alpha$

Component of velocity $v_y$ will be zero when the ball reaches the hoop because there, the velocity is parallel to inclined plane.
$v_y=u_y+a_{y}t$
$\therefore 0=u\sin \alpha +(-g\cos \theta )t$
$\Rightarrow u\sin \alpha =g\cos \theta .t...\left(1\right)$
Given: displacement of ball in $y$ direction is $s_y=4\text{m}$
$v_y^2=u_y^2+2a_y{s}_y$
$\Rightarrow 0=(u\sin \alpha {)}^2+2\times (-g\cos \theta )\times 4....(2)$
From eq. $\left(1\right)\&\left(2\right)$,
$(g\cos \theta \times t{)}^2=2\times g\cos \theta \times 4$
or, $g\cos \theta \times t^2=2\times 4$
$\Rightarrow t^2=\frac{2\times 4}{10\times \frac{4}{5}}$
$\therefore t=1\text{second}$