Question

A marble is dropped off the top of a building. Find it's velocity and how far below to its dropping point for the first 10 seconds of the drop .Use g=-9.8m/s^2. Please see the chart below and help me if you can. Thank you .

Answer 1

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