2points
an athlete jumps with a speed of 4 m/s at an angle of 23º. how long does the
athlete stay in the air?
o
a. 0.32 s
b. 0.92 s
c. 0.55 s
d. 0.75 s
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Calculate the vertical component of the athlete's speed:

Vy = Vsin(θ)

Vy is the vertical component of the speed, V is the speed, and θ is the angle the athlete jumps off the ground at.

Given values:

V = 4m/s

θ = 23°

Plug in these values and solve for Vy:

Vy = 4sin(23°) = 1.563m/s

Let us use this kinematics equation for the athlete's vertical motion:

H = Vt + 0.5At²

H is the height, t is time, V is the initial vertical velocity, and A is acceleration.

Given values:

V = 1.563m/s

A = -9.81m/s² (acceleration due to gravity)

Plug in these values:

H = 1.563t - 4.905t²

We want to know the athlete's airtime, or when they reach the ground, ie calculate a time t when H = 0m. So let us substitute H = 0 and solve for t:

1.563t - 4.905t² = 0

t = 0.3187, 0

Reject t = 0s

t = 0.3187s

Choice A

the answer is b i think

it doesn't. it was imagined up in the middle ages.