Along jumper jumps at a 20 degree angle and attains a maximum altitude of 0.6 m

Refer to the diagram below.

Assume g = 9.8 m/s² and ignore air resistance.

Let V =  the launch velocity.
The horizontal component of the launch velocity is
Vx = V cos(20°) = 0.9397V m/s
The vertical component of the launch velocity is
Vy = V sin(20°) = 0.342V m/s

Let t =  the time to attain the maximum height of 0.6 m.
At maximum height, the vertical velocity is zero, therefore
Vy - 9.8 t = 0
Vy = 9.8t                   (1)

Also,
Vy*t - (1/2)*9.8*t² = 0.6               (2)

Substitute (1) into (2).
(9.8t)*t - 0.5*9.8*t² = 0.6
4.9t² = 0.6
t = 0.35 s

Therefore,
Vy = 9.8*0.35 = 3.43 m/s
V = Vy/0.342 = 10.0292 m/s
Vx = 0.9397V = 9.4244 m/s

The total time of travel is 2t = 2*0.35 = 0.7 s.
The horizontal travel is
d = 9.4244 *0.7 = 6.597 m

Answers:
The time for the jump is 0.7 s.
The length of the jump is 6.6 m (nearest tenth)

As newton's first law states a object in motion stays in motion

The answer should be a.