A long 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)