A pebble is dropped from rest from the top of a till cliff and falls `4.9 m` after `1.0 s` has elapsed. How much farther does it drop in the next `2.0 s` ? `("Take" g - 9.8 m//s^2)`.

Options

A. 9.8
B. 19.6 m
C. 39 m
D. 44 m

Answer

C

Explanation

We take downward as the positive direction with `y = 0 and t = 0` at the top of the cliff. The freely falling pebble then has `v_0 = 0 and a = g = +9.8 m//s^2`. The displacement of the pebble at `t = 1.0 s` is given by `y_1 = 4.9 m`. The displacement of the pebble at `t = 3.0 s` is found from.
`y_3 = v_0 t + (1)/(2) at^2 = 0 + (1)/(2) (9.80 m//s^2)(3.0 s)^2 = 44 m`
The distance fallen in the `2.0 -s` interval from `t = 1.0 s` to `t = 3.0 s` is then
`Delta y = y_3 - y_1 = 44 m - 4.9 m = 39 m`.