Question

The small marble is projected with a velocity of $10m{s}^{-1}$ in a direction ${45}^0$ from the horizontal y-direction on the smooth inclined plane. Calculate the magnitude v of its velocity after $2$ s.

• A. $10\sqrt{2}m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $10m{s}^{-1}$
• D. $5\sqrt{2}m{s}^{-1}$

Answer 1

The correct option is C $10m{s}^{-1}$

Final velocity in vertical direction :-

$V_y=V_y+a_{y}t$

$=10\sin 45-g\cos 45\times t$

$=\frac{10}{\sqrt{2}}-\frac{10}{\sqrt{2}}\times 2$

$=\frac{-10}{\sqrt{2}}\hat{j}\frac{m}{s}$

Horizontal velocity = $V_x$

$V_x=U_x-a_{x}t$

$10\cos 45-g\sin 45\times 2$

$=\frac{-10}{\sqrt{2}}\hat{i}m{s}^{-1}$

net velocity$=\sqrt{\frac{100}{2}+\frac{100}{2}}=\frac{10m}{s}$