consider the motion in y-direction
v₀ = initial velocity = 0 m/s
a = acceleration = g = - 9.8 m/s²
t = time
v = final velocity at any time "t"
velocity at any time is given as
v = v₀ + at
v = 0 + (- 9.8) t
v = (- 9.8) t
so at t = 1 , v = (- 9.8) (1) = - 9.8 m/s
at t = 2 , v = (- 9.8) (2) = - 19.6 m/s
at t = 3 , v = (- 9.8) (3) = - 29.4 m/s and so on
Y₀ = initial position where the marble was dropped from = 0 m
Y = final position at any time "t"
final position at any time "t" is given as
Y = Y₀ + v₀ t + (0.5) a t²
Y = 0 + (0) t + (0.5) (-9.8) t²
Y = - 4.9 t²
at t = 1 , Y = - 4.9 (1)² = - 4.9 m
at t = 2 , Y = - 4.9 (2)² = - 19.6 m
at t = 3 , Y = - 4.9 (3)² = - 44.1 m
and So on
