A boy of height `1.5 m`, making move on a skateboard due east with velocity `4 m s^-1`, throws a coin vertically up a velocity of `3 m s^-1` relative to himself.
(a) Find the total displacement of the coin relative to ground till it comes to the hand of the boy.
(b) What isthe maximum height attained by the coin w.r.t. to ground ?

Explanation

The path of the coin as seen from boy `A` is a straight line (up and down). But that seen from ground observer `B` is parabola.
Velocity of coin, `vec v_("coin") = vec v_("coin, boy") + vec v_("boy") vec v_("coin") = 4 hat i + 3 hat k(m s^-1)`
Acceleration of coin, `vec a_("coin") = -10 hat k(m s^-2)`
The displacement of coin in z - direction will be zero.
Using `vec s = vec u t + (1)/(2) xx 10 xx (T)^2` (T = Time of flight of coin)
or `0 = T(3 - 5T) rArr T = (3)/(5) s`
The displacement of the coin will be only x - direction.
Let the displacement in x - direction be `x`. Then
`x = v_x xx T = 4 xx (3)/(5) = (12)/(5) m`
At maximum height, the final component of coin velocity in vertical direction will be zero.
Using `v^2 = u^2 + 2` as in vertical direction,
`0 = (3)^2 - 2.10 H rArr H = (9)/(20) m`
Hence, maximum height reached by coin
`H_(max) = Height of boy + H = 1.5 + (9)/(20) = (39)/(20) m`
(a) Hence the displacement of coin will be in horizontal direction and will be equal to `12//5 m`.
(b) The maximum height attained by coin will be `39//20 m`.
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