Chapter 3, Problem 98A

(a)

To determine

The velocity of the bag attained after 2.0s if it is dropped from a helicopter.

Answer to Problem 98A

  $14.6\text{m/s}$

Explanation of Solution

Given:

A bag of cargo is dropped from a helicopter which is rising at $5.0\text{m/s}$ .

Formula used:

Final velocity with constant acceleration is given by,

  $v_f=v_i+a\Delta t$

Where, $v_f$ is the final velocity, $v_i$ is the initial velocity, a is the acceleration and $\Delta t$ is the time taken.

Calculation:

Assume the downward direction as positive and upwards direction as negative.

Now, since the bag is dropped from helicopter, thus its initial velocity will be that of the helicopter, that is,

  $v_i=-5\text{m/s}$

Acceleration acting on it will be due to gravity only. Thus,

  $a=9.8{\text{m/s}}^{\text{2}}$

Then, using the above-mentioned formula for the final velocity, the velocity of the bag after $2.0\text{ s}$ will be

  $\begin{array}{c}{v}_f=v_i+a\Delta t\\ =-5+\left(9.8\times 2.0\right)\\ =+14.6\text{m/s}\end{array}$

Positive sign denotes the downward direction.

Conclusion:

Thus, the velocity of the bag after 2.0s is $14.6\text{m/s}$ .

(b)

To determine

The distance by which the bag has fallen in 2.0s.

Answer to Problem 98A

  $9.6\text{m}$

Explanation of Solution

Given:

A bag of cargo is dropped from a helicopter which is rising at $5.0\text{m/s}$ .

Formula used:

The distance traveled by an object with a constant acceleration is given by:

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

Where $d_f$ is the final distance covered, $d_i$ is the initial distance of the object, $v_i$ is the initial velocity, a is the acceleration and $\Delta t$ is the time taken.

Calculation:

Assume the downward direction as positive and upwards direction as negative.

Now, since the bag is dropped from helicopter, thus its initial velocity will be that of the helicopter, that is,

  $v_i=-5\text{m/s}$

And, only acceleration acting on it will be due to gravity only. Thus,

  $a=9.8{\text{m/s}}^{\text{2}}$

Then, using the above mentioned formula to determine the distance by which the bag is fallen as,

  $\begin{array}{c}{d}_f=d_i+v_i\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ =0+\left(-5.0\right)\left(2.0\right)+\frac{1}{2}\left(9.8\right){\left(2.0\right)}^2\\ =9.6\text{m}\end{array}$

Conclusion:

Thus, the distance by which the bag has fallen in 2.0s is $9.6\text{m}$ .

(c)

To determine

The distance between the bag and the helicopter when the bag has fallen for 2.0s.

Answer to Problem 98A

  $19.6\text{m}$

Explanation of Solution

Given:

A bag of cargo is dropped from a helicopter which is rising at $5.0\text{m/s}$ .

Formula used:

The distance traveled by an object with a constant acceleration is given by,

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

Where $d_f$ is the final distance covered, $d_i$ is the initial distance of the object, $v_i$ is the initial velocity, a is the acceleration and $\Delta t$ is the time taken.

Calculation:

Assume the downward direction as positive and upwards direction as negative.

Now, since the bag is dropped from helicopter, thus its initial velocity will be that of the helicopter, that is,

  $v_i=-5\text{m/s}$

And, only acceleration acting on it will be due to gravity only. Thus,

  $a=9.8{\text{m/s}}^{\text{2}}$

Then, using the above mentioned formula to determine the distance by which the bag is fallen as,

  $\begin{array}{c}{d}_f=d_i+v_i\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ =0+\left(-5.0\right)\left(2.0\right)+\frac{1}{2}\left(9.8\right){\left(2.0\right)}^2\\ =9.6\text{m}\end{array}$

The height by which the helicopter has risen in 2.0s is,

  $\begin{array}{c}d\text{'}=\left|\text{speed}\times \text{time}\right|\\ =\left|\left(-5\right)\times 2\right|\\ =10.0\text{m}\end{array}$

Then, the total distance between bag and the helicopter after 2.0s is,

  $\begin{array}{c}{d}_t=d_i+d\text{'}\\ =\left(9.6+10\right)\\ =19.6\text{m}\end{array}$

Conclusion:

Thus, the distance between the bag and the helicopter after 2.0s is $19.6\text{m}$ .