Where have I gone wrong? A basic question in physics (Mechanics) A stone is dropped vertically in a velocity of $15$ minutes per second, from a point $40$ meters above the ground (excuse my poor English.).
a. How long will it take to the stone to hit the ground?
I answered it like this: 
$v_0=15$, $a(t)=a_0=g$, $v(t)=v_0+gt$ $\Rightarrow x(t)=x_0+v_0t+{1\over 2}gt^2\Rightarrow x(t)=x_0+15t+gt^2$.
I denoted $x_0=0$, which derived $x(t_1)=40$ given $t_1$ is the time it takes to get to the ground. 
Then I solve $x(t_1)=40=15t_1+4.9t_1^2$. I got that $t_1=1.7s$ sec, and the other option is negative. 
The problem is: I was then asked: when will the stone be at 2 sec? 3 sec? That makes me feel like I got this all wrong. I am new to this an I am not sure I know what I am doing. I would appreciate your help.  
 A: The solution is no more complex. You have to use the equation
$y(t)=h_0+v_0t-\frac{1}{2}gt^2$, where $h_0$ is the initial position ($40$m), and $v_0$ the initial velocity ($0$ in this case). Now, solve
$40-4.9t^2=0$
and use the original fórmula for the other questions too.
A: Hint:
You have a problem in two dimension with:
$ \vec v_0=(v_0,0)^T$ (and I suppose your velocity is in meter/sec)
$\vec a=(0,-g)^T$
$\vec s_0=(40,0)$
Your equation of motion become:
$$
\vec s(t)=\dfrac{t^2}{2}\vec a+t \vec v_0+\vec s_0 \Rightarrow
\begin{bmatrix}x(t)\\y(t) \end {bmatrix}=\dfrac{t^2}{2}\begin{bmatrix}0\\-g \end {bmatrix}+t\begin{bmatrix}15\\0 \end {bmatrix}+\begin{bmatrix}0\\40 \end {bmatrix}
$$
Solve the $y$ componet for $y(t)=0$ and you have the time to hit the ground, use this time in the $x$ componet and you find the point where the stone stops.
A: Don't feel bad.I think the wording of the problem is incorrect.If I had to guess,it would go something like this:"A stone is dropped vertically from a height of 40 meters above the ground at an initial velocity of 15 meters/sec.
We need to measure distance from 40 meters up toward the ground,If s = distance,
s = 0 at 40 meters up.The acceleration of gravity is about 9.8 meters/(sec^2)'
We start with the differential equation: d2/s2 = 9.8.(pardon the notation for
the second derivative.)Integrating ds/dt = 9.8t + v0.or ds/dt = 9.8t + 15.We
integrate again to get a formula for the distance,s. s = 4.9t^2 + 15t + s(0).
But s(0) = 0 and s = 40.This gives a quadratic equation in t=4.9t^2 + 15t - 40=0
Solve for t using the quadratic formula:9.8t=-15+sqrt(225+784),or t = 1.71 sec.
I think the other parts mean to say:"WHERE IS THE STONE AT 2 SEC.,3 SEC., ETC.
Just plug the values into your equation for s and you will find the stone will
hit the ground before these times can be reached.   Edwin Gray
