Sec(360)=1/cos(360). 

If we take a stick of length l and fix one of its end to the origin. It is free to rotate from its another end and will make circle at Cartesian coordinate plane. Rotate it by 90º it will lie on positive y axis. Similarly rotate it by 180° it will lie on negative x axis. If we rotate it by 360° we will find that it came in its original position i.e lies on positive x axis. Imagine a position where stick was just about complete. Take the projection of stick on x axis. We get a right angle triangle in third quadrant. Calling the angle between positive x axis and stick as θ. Now cosθ=(projection of stick on positive x axis) / (length of stick=l). 

As θ decreases. Projection of stick will increase and when stick completes its 360° rotation it will be equal to l. 

At that time 

Cosθ = l/l =1.

⇒secθ = 1/cosθ =1/1 =1 Ans. 

Total time taken by the particle to go up and come back = 5+9 =14s.

Therefore it must take 7s to go up and 7s to come down. 

From first law, we know

v=u+at

At the top point velocity will be zero. Therefore v=0.

0=u-10*7

(taking a=g=-10m/s²,  negative sign indicates that acceleration is in downward direction). Solving the equation we will get

u=70m/s.

It is given that after 5s its height was h. 

Again applying first law. 

v=u+at

v=70-10*5

v= 20m/s

Particle speed was 20m/s in upward direction at height h after 5s of throwing.