Question

Two balls are thrown from an inclined plane at angle of projection $\alpha$ with the plane, one up the incline and other down the incline as shown in figure (t stands for total time of flight):

• A.
• B.
• C.
• D.

Answer 1

Answer: (A), (B), (C)

The correct options are
A

B

C

Range over the incline is the distance travelled along the plane in the given time period.

$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$ $S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$R_1=v_{o}cos\alpha T-\frac{1}{2}gsin\theta T^2$ $-R_2=v_{o}cos\alpha T-\frac{1}{2}gsin\theta T^2$

$R_2-R_1=gsin\theta T^2$

So option C is correct

$v_{t_1}\&v_{t_2}$ are the velocities of the particles at their maximum heights. Let the particles reach their maximum heights at time $t_1$ and $t_2$ respectively. Hence

$\Rightarrow v_y=u_y+a_{y}t$

$0=\left(v_{0}sin\alpha \right)-\left(gcos\theta \right)t_1\Rightarrow t_1=\frac{v_{o}sin\alpha }{gcos\theta }$

Similarly $t_2=\frac{v_{o}sin\alpha }{gcos\theta }$ Hence $t_2=t_1$

Hence $v_{t_1}=v_{o}cos\alpha -\left(gsin\theta \right)t_1$

$v_{t_2}=(-v_{o}cos\alpha )-\left(gsin\theta \right)t_2\Rightarrow v_{t_1}\ne v_{t_2}$

Range over the incline is the distance travelled along the plane in the given time period.

$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$ $S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$R_1=v_{0}cos\alpha T-\frac{1}{2}gsin\theta T^2$ $-R_2=v_{o}cos\alpha T-\frac{1}{2}gsin\theta T^2$

$R_2-R_1=gsin\theta T^2$

So option C is correct

$v_{t_1}\&v_{t_2}$ are the velocities of the particles at their maximum heights. Let the particles reach their maximum heights at time $t_1$ and $t_2$ respectively. Hence

$\Rightarrow v_y=u_y+a_{y}t$

$0=\left(v_{o}sin\alpha \right)-\left(gcos\theta \right)t_1\Rightarrow t_1=\frac{v_{o}sin\alpha }{gcos\theta }$

Similarly $t_2=\frac{v_{o}sin\alpha }{gcos\theta }$ Hence $t_2=t_1$

Hence $v_{t_1}=v_{o}cos\alpha -\left(gsin\theta \right)t_1$

$v_{t_2}=(-v_{o}cos\alpha )-\left(gsin\theta \right)t_2\Rightarrow v_{t_1}\ne v_{t_2}$