AS integration

question: A stone is thrown vertically upwards from ground level with a velocity of 12.6m/s. If the acceleration due to gravity is $9.8m/s^2$, deduce from the differential equation =$\frac{dv}{dt}=-9.8$, expressions for its velocity and its height t s later.
Find:

a)the time to the highest point
b)the greatest height reached,
c)the distance moved through by the stone during each of the first two seconds

Answer:
a) a = -9.8 $v = \int (-9.8)dt = \left [ -9.8t + c\right ] = -9.8t +c$

(0, 12.6)
therefore
$v =12.6=-9.8(0)+c$
thus; $v=-9.8t +12.6$
v =0
[text] 12.6-9.8t =0