A level Mechanics 1

A small ball is projected vertically upwards with speed 29.4ms-1 from a point A which is
19.6m above horizontal ground.
The ball is modelled as a particle moving freely under gravity until it hits the ground. It is
assumed that the ball does not rebound.
(a) Find the distance travelled by the ball while its speed is less than 14.7ms-1
(3)
(b) Find the time for which the ball is moving with a speed of more than 29.4ms-1
I could solve no-a but I'm tearing my head trying to solve to no-b while assigning sign conventions.I took upwards as positive,so shouldn't be equation come 29.4t-4.9t2=-19.6? If I'm wrong,please do correct me,and please explain the appropriate sign conventions to be used here

A sketch is always useful, but it should be fairly easy to reason about when the speed is equal to 29.4 (in relation to the initial point), and so the section when the speed is > 29.4 should be clear. Youll need that time and the total time of flight.

Okay but how do we use s=ut+1/2at2 here with appropriate sign conventions?

The only thing youd use that for is the total time of flight which you had correct in the OP, so if up is positive a=-9.8, u=29.4 and s=-19.6.

I got t=6.61s (2dp) and t= -0.61s,but in the mark scheme the answers are to=6.61 and t=0.61s,they used 19.6=29.4t+4.9t2