A stone is thrown vertically upward with a speed of 18.0 m/s.

Question

(a)How fast is it moving when it reaches a height of 11.0 m?
(b)How long is required to reach this height ?
(c)Why are there two answers to (b)?

Answer 1

first c
It passes 11 meters on the way up and on the way down.

now the calculations

I guess it started at h = 0 (we are lying on our backs in a shallow hole throwing this upward )

Vi = 18
a = -9.81 m/s^2 = g

initial Ke = (1/2) m v^2 = 162 m
loss of Ke at 11 meters = m g h
= 108 m
remaining Ke at 11 meters = (162-108)m = 54 m
so
(1/2) m v^2 = 54 m
v = 10.4 m/s on the way up, -10.4 m/s on the way down by symmetry(part a)

part b
v = Vi - 9.81 t
10.4 = 18 - 9.81 t
t = .775 seconds on the way up
or
-10.4 = 18 - 9.81 t
- 28.4 = -9.81 t
t = 2.9 seconds on the way down