For (c), at max height, v=0, so use v2 = u2 + 2as, just put u=20, a=-9.81, s=54

[madinsane]

Homework Statement

a man throws a stone vertically upward with initial speed of 20 m/s from a building with height 54 m a)how long does the stone take to reach 40 m/s
b)what is its velocity before it reaches the ground
c) what is the magnitude of its acceleration at its maximum height?

Homework Equations

vf=vi+at
xf=xi+vit+0.5at
vf^2=vi^2+2a(xf-xi)

The Attempt at a Solution

I just want someone to tell me whether what I have done is right or wrong
so for a)
I took vi=20m/s
vf=0m/s (I was going to calculate how long it takes to reach the maximum peak)
then I used vf=vi+at (a as-9.8)
I got 2.04 s
Then I took vi=0
vf=40
then used the same thing and got t as ~4 and then I added them together
b) At first I calculated the distance it takes to go from the building to the peak
using xf=xi+vit+0.5at
xi=0
vi=20
t=2.04
a=-9.8
I got ~22m
Then I added to the 54
then I used vf^2=vi^2+2a(xf-xi)
vi=0
a=-9.8
xf=76
xi=0
and got vfinal
c) I said it was 9.8 m/s^2
So did I do anything wrong and if I did tell me what it is

 

[th4450]

You're correct, but it can be done simpler
For (a), v = u + at, just put v=-40, u=20, a=-9.81
For (b), v² = u² + 2as, just put u=20, a=-9.81, s=-54