The small marble is projected with a velocity of 10 m/s in a direction 45o from the horizontal y-direction on the smooth inclined plane. Calculate the magnitude v of its velocity after 2 s.

10 m/s

Given : v = 10 m/s

After 2 s : vx = \(\frac{10}{\sqrt{2}} - \frac{10}{\sqrt{2}} x 2\)

v_x = - \(\frac{10}{\sqrt{2}}\) and v_y = - \(\frac{10}{\sqrt{2}}\)

v = \(\sqrt{(\frac{10}{\sqrt{2}})^2 + (\frac{10}{\sqrt{2}})^2}\)

v = 10 m/s