A small sphere is projected with a velocity of `3 ms^-1` in a direction `60^@` from the horizontal y - axis, on the smooth inclined plane (Fig. 5.197). The motion of sphere takes place in the x - y plane. Calculate the magnitude `v` of its velocity after `2 s`.
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Answer

`1.5 m s^-1`

Explanation

The motion of the sphere takes place in a plane , the x - and x - components of its acceleration are `A_x = -g sin 30^@`, `a_y = 0` The x- and y - components of the sphere's velocity at time `t = 2 s` are
`V_x = v_(0 x) - a_x t =3 sin 60^@ - g sin 30^@ xx 2 = 0.1 ms^-1`
`V_y = v cos 60^@ = constant = 1.5 ms^-1`
So the magnitude of sphere's velocity is
`|vec v| = sqrt(v_x^2 + v_y^2) = sqrt((0.1)^2 + (1.5)^2) ~= 1.5 ms^-1`
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