91Ó°ÊÓ

In this problem, the downward direction is considered positive, and the origin is taken to be located at the plane's location.

Given data:

The initial altitude of free fall is\({d_1} = 3200\;{\rm{m}}\).

The final altitude of free fall is\({d_2} = 450\;{\rm{m}}\).

The speed of the skydiver is\(v = 150\;{\rm{km/h}}\).

The difference in distances is calculated as

\(\begin{aligned}y &= {d_1} - {d_2}\\y &= 3200\;{\rm{m}} - 450\;{\rm{m}}\\y &= {\rm{2750}}\;{\rm{m}}\end{aligned}\)

The relation from the kinematics equation is given by

\(y = ut + \frac{1}{2}g{t^2}\)

Here, g is the gravitational acceleration, and u is the initial speed of the parachutist whose value is zero.

On plugging the values in the above relation, you get:

\(\begin{aligned}2750\;{\rm{m}} &= 0 + \frac{1}{2}\left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right){\left( t \right)^2}\\t &= 23.6\;{\rm{s}}\end{aligned}\)

The relation from the kinematics equation is given by:

\(v = u + gt\)

On plugging the values in the above relation, you get:

\(\begin{aligned}v &= 0 + \left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right)\left( {23.6\;{\rm{s}}} \right)\\v &= 231.5\;{\rm{m/s}}\end{aligned}\)

Thus, \(v = 231.5\;{\rm{m/s}}\) is the required speed and time is \(t = 23.6\;{\rm{s}}\).