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.
Answer & explanation
Correct answer: option 3
Given : v = 10 m/s
After 2 s : vx = \(\frac{10}{\sqrt{2}} - \frac{10}{\sqrt{2}} x 2\)
vx = - \(\frac{10}{\sqrt{2}}\) and vy = - \(\frac{10}{\sqrt{2}}\)
v = \(\sqrt{(\frac{10}{\sqrt{2}})^2 + (\frac{10}{\sqrt{2}})^2}\)
v = 10 m/s