Answer is: option1

Solution:

Calculation of Maximum Upward Velocity

The velocity function is the first derivative of height \( h(t) \):

\[ v(t) = h'(t) = \frac{d}{dt} \left( t^3 - 6t^2 + 20t \right) \]

\[ v(t) = 3t^2 - 12t + 20 \]

The velocity is maximum when its derivative (acceleration) is zero. Find the acceleration function by differentiating \( v(t) \):

\[ a(t) = v'(t) = \frac{d}{dt} (3t^2 - 12t + 20) \]

\[ a(t) = 6t - 12 \]

Set acceleration to zero:

\[ 6t - 12 = 0 \]

\[ t = 2 \]

Substitute \( t = 2 \) into the height function:

\[ h(2) = (2)^3 - 6(2)^2 + 20(2) \]

\[ = 8 - 6(4) + 40 \]

\[ = 8 - 24 + 40 \]

\[ = 24 \]

The height at the instant the object reaches its maximum upward velocity is 24 meters, which corresponds to option (A).