Question

A stone is thrown vertically upward with a speed of 11.0 m/s from the edge of a cliff 66.0 m high.

(a) How much later does it reach the bottom of the cliff?
s
(b) What is its speed just before hitting?
m/s
(c) What total distance did it travel?
m

Answer 1

use equations of constant acceleration
displacement s = -66, u = 11, a = -9.8, t = ?
s = ut + 0.5at^2
-66 = 11t - 4.9t^2
4.9t^2 - 11t - 66 = 0
t = 4.96 secs

Answer 2

find time to reach greatest height
v = 0, u = 11, a = -9.8, t =?
o = 11 + -9.8 * t
t = -11/-9.8 = 1.12 secs

find v when stone hits ground:
u = 0, t = 4.96-1.12 = 3.84, v = ?, a=9.8
v = 0 +9.8 * 3.84 = 37.63 m/s

Answer 3

distance travelled ascending
u=11, a=9.8,s =?,v = 0
v^2 = u^2 -2*9.8 * s
0 = 121 - 19.6 s
s = 121/19.6 = 6.173m
distance travelled when passing intitial postion = 2 * 6.173 = 12.35m

total distance = 12.35 + 66 = 78.35 m

Answer 4

thank you thank you thank you thank you.
could you help me with my other problem too?

Answer 5

i'll try