Question

A ball is projected on smooth inclined plane in direction perpendicular to line of greatest slope with velocity of $8\text{m/s}$ .Find its speed after $1\text{sec}$ (take $g=10{\text{m/s}}^2$ ).

• A. $8\text{m/s}$
• B. $12\text{m/s}$
• C. $10\text{m/s}$
• D. $14\text{m/s}$

Answer 1

The correct option is C $10\text{m/s}$


Given the ball is projected perpendicular to the line of greatest slope . So, the ball is projected on the inclined plane parallel to the horizontal ($x-$ axis).

So, there is no acceleration in $\text{x-}$direction and hence , $u_x=8\text{m/s}$ remains constant in this direction .

But the plane is inclined with the $\text{y-}$axis at some angle. So, the component of acceleration downwards the incline plane is given by,

$a_y=g\sin {37}^{\circ }=10\times \frac{3}{5}=6{\text{m/s}}^2$

Initial velocity along the $\text{y-}$axis is given by, $u_y=0$

Now, velocity along the $\text{y-}$axis at time, $t=1\text{sec}$ can be calculated by using equation of motion,

$v_y=u_y+a_{y}t$

Substituting the values, we get

$v_y=0+6\times 1=6\text{m/s}$

So, the net speed after $1\text{sec}$ will be

$v_{net}=\sqrt{u_x^2+v_y^2}$

$\Rightarrow v_{net}=\sqrt{8^2+6^2}=10\text{m/s}$

Hence, option (c) is the correct answer.