Chapter 3, Problem 73A

To determine

The final velocity and distance covered by the bag when it has fallen for 2.0s.

Answer to Problem 73A

  $v=19.6\text{m/s}$

  $d=19.6\text{m}$

Explanation of Solution

Given:

A bag is dropped from a helicopter and has fallen for 2.0s.

Formula used:

Final velocity with constant acceleration can be obtained by:

  $v_f=v_i+a\Delta t$

Where, $\Delta t$ is the time taken, a is acceleration and v represents velocity.

The distance traveled by an object with a constant acceleration can be obtained by:

  $d_f=d_i+v_i\Delta t+\frac{1}{2}a\Delta t^2$

Here, d represents displacement, v represents velocity, t represents time and a is the acceleration.

Calculation:

Since, the bag is dropped from a helicopter, thus its initial velocity will be 0, that is,

  $v_i=0\text{m/s}$

Acceleration acting on it will be due to gravity only (as air resistance is to be ignored). Thus,

  $a=9.8{\text{m/s}}^{\text{2}}$

Now, using the above-mentioned formula for the velocity, the final velocity of the bag after 2.0s will be:

  $\begin{array}{c}{v}_f=v_i+a\Delta t\\ =0+\left(9.8\right)\left(2.0\right)\\ =19.6\text{m/s}\end{array}$

Assuming the initial distance to be 0, then

  $d_i=0\text{}\text{m}$

Then, the distance covered in the time 2.0 s will be,

  $\begin{array}{l}{d}_f=d_i+v_i\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ =0+0\left(2\right)+\frac{1}{2}\left(9.8\right){\left(2\right)}^2\\ =19.6\text{}\text{m}\end{array}$

Conclusion:

Thus, the final velocity of the bag and distance covered by the bag in the given time is $19.6\text{m/s }$ and $19.6\text{ m}$ .