velocity/accerleration of ball question, see if im doing this right.?

Question

i got answers for this question yesterday, i just want to make sure I can solve them right so tell me if what im doing seems correct. And if you correct me try to solve it using one of these formulas:

V=at+V0
(X-X0)= 1/2 (V+V0)t
V^2=V0^2 + 2a (X-X0)
X-X0 = 1/2at^2 + V0t

now...check and see if what I say sounds good

A ball is thrown vertically upward with a velocity of +10.0 m/s.
(a) How high does it rise?
___m

Using V^2=V0^2 + 2a (X-X0)
10^2=0^2 + 2 * 9.8 *x, x = 5.10m

(b) How long does it take to reach its highest point?
___s

Using V=at+V0

10=9.8x, x= 1.0204sec

(c) How long does the ball take to hit the ground after it reaches its highest point?
___s

Heres where I get a bit confused, which equation should I use for this? Would the answer be the exact same? or would it be different a little? I know its pretty much 1 sec

(d) What is its velocity when it returns to the level from which it started?
___m/s

Also help with this one please, I think its just 9.8, thanks

Answer 1

Question a)
v² = u² - 2gs (let g = 10)
0 = 100 - 20s
s= 5
It rises 5m

Question b)
v = u - gt
0 = 10 - 10t
t = 1 sec

Question c)
s = ut + (1/2) gt²
5 = 1/2 x 10 x t²
t² = 5/5
t = 1 sec

Question d)
v = 0 + g x 1
v = 10 m / s

Answer 2

The first one is okay but the units look messy. I did not check the others. I strongly suggest you look at units (called dimensional analysis.)

Going over #1

I suggest you use JUST the formulae
X-X0 = V0 + 0.5 aT^2
and
V = V0 + aT.

In this problem the ball rises until V = 0. This happens when V0 = AT. So using 10 for a (approximately true) you get T = 1.

Now you use the in the formula for X.

Answer 3

The fall time from peak will be the same a rise time from initial point.

The velocity when at the same level as at start is the same as the initial velocity. (The kinetic energy must be the same at end as at start,)