Verified answer

(15 m/s)/(9.8 m/s^2) = about 1.5 seconds.

At the top of its motion, the vertical speed is zero. So the starting speed of 15 m/s goes to zero. The projectile speed goes down by 9.8 m/s, each second. So how long does it take to reduce 15 m/s to 0, at that rate?

at the top of its motion the projectile will have a velocity of 0 (i.e. it "stalls" out and begins to fall back down)

v = u + at

0 = 15 - 9.8t

t = 1.5 s