Chapter 2, Problem 80GP

To determine

The time taken by a parachutist. Also, the speed at the given distance.

Answer to Problem 80GP

  $24.12\sec$

  $236.4\frac{\text{m}}{\text{s}}$

Explanation of Solution

Given:

Initial velocity of the parachutist as it falls from the plane $=v_o=0\frac{\text{m}}{\text{s}}$

Initial position of the parachutist as it falls from the plane $=y_o=3200\text{m}$

Final position of the parachutist $=y=350\text{m}$

Acceleration of the parachutist $=a=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Time of travel $=t$

Final velocity of the parachutist as she pulls her ripcord $=v_f$

Acceleration of the parachutist $=a=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Formula Used:

Position of an object at any time in accelerated motion is given as

  $y=y_o+v_{o}t+(0.5)a{t}^2$

Velocity at any time for an object in accelerated motion is given as

  $v_f=v_o+at$

Calculation:

Final position of the parachutist is given as

  $y=y_o+v_{o}t+(0.5)a{t}^2$

Inserting the values

  $\begin{array}{l}350=3200+(0)t+(0.5)(-9.8)t^2\\ -2850=-4.9t^2\\ t=24.12s\end{array}$

Velocity of the parachutist is given as

  $v_f=v_o+at$

Inserting the values

  $\begin{array}{l}{v}_f=0+(-9.8)(24.12)\\ v_f=-236.4\frac{\text{m}}{\text{s}}\end{array}$

Negative sign indicates opposite direction.

Conclusion:

Therefore, the time taken is 24.12 sec and speed is $236.4\frac{\text{m}}{\text{s}}$