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