We have that the instantaneous velocity of the
shuttlecock when it hits the ground is

From the question we are told
Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
shuttlecock when it hits the ground? Show your work below.
Generally the equation for acceleration is mathematically given as

Where
acceleration is still -9.81 m/s2,
Hence,

Therefore

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Answer:
t_total = 23.757 s
Explanation:
This is a kinematics exercise.
Let's start by calculating the distance and has to reach the limit speed of
v = 18.8 m / s
v = v₀ + a t₁
the elevator starts with zero speed
v = a t₁
t₁ = v / a
t₁ = 18.8 / 2.40
t₁ = 7.833 s
in this time he runs
y₁ = v₀ t₁ + ½ a t₁²
y₁ = ½ a t₁²
y₁ = ½ 2.40 7.833²
y₁ = 73.627 m
This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.
x_total = x₁ + x₂
x₂ = x_total - x₁
x₂ = 373 - 73,627
x₂ = 299.373 m
this distance travels at constant speed,
v = x₂ / t₂
t₂ = x₂ / v
t₂ = 299.373 / 18.8
t₂ = 15.92 s
therefore the total travel time is
t_total = t₁ + t₂
t_total = 7.833 + 15.92
t_total = 23.757 s