Chapter 3, Problem 76A

(a)

To determine

The height attained by the beanbag in the air.

Answer to Problem 76A

  $5.929\text{m}$

Explanation of Solution

Given:

A beanbag is thrown in the air and it took $2.2\text{ s}$ to reach its initial position.

Formula used:

Final velocity with constant acceleration can be obtained by:

  $v_{final}=v_{initial}+a\Delta t$

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:

  $d_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a\Delta t^2$

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

Calculation:

Any object comes at momentarily rest at maximum height, thus the velocity of the beanbag at the maximum height will be 0. That is,

  $v_f=0$

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

  $a=9.8{\text{m/s}}^{\text{2}}$

Also, since the motion of the bag will be symmetric since uniform acceleration acts on it, thus time taken by bag to reach maximum height will be

  $t=\frac{2.2}{2}=1.1\text{s}$

Considering the upward motion of the bag, acceleration will be $a=-9.8{\text{m/s}}^{\text{2}}$ (negative sign denotes downward direction). Using the above-mentioned formula for the velocity, the initial velocity of the bag can be calculated as

  $\begin{array}{c}{v}_f=v_i+a\Delta t\\ 0=v_i+\left(-9.8\right)\left(1.1\right)\\ v_i=10.78\text{m/s}\end{array}$

Then, the maximum height attained in 1.1s will be:

  $\begin{array}{c}{d}_f=d_i+v_i\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ =0+10.78\left(1.1\right)+\frac{1}{2}\left(-9.8\right){\left(1.1\right)}^2\\ =\left(11.858-5.929\right)\text{m}\\ =5.929\text{m}\end{array}$

Conclusion:

Thus, the maximum height attained bythe beanbag is $5.929\text{m}$ .

(b)

To determine

The initial velocity of the beanbag with which it was thrown in the air.

Answer to Problem 76A

  $10.78\text{m/s}$

Explanation of Solution

Given:

A beanbag is thrown in the air and it took 2.2s to reach its initial position.

Formula used:

Final velocity with constant acceleration is given by,

  $v_f=v_i+a\Delta t$

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

Calculation:

Any object comes at momentarily rest at maximum height, thus the velocity of the beanbag at the maximum height will be 0. That is,

  $v_f=0$

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

  $a=9.8{\text{m/s}}^{\text{2}}$

Also, since the motion of the bag will be symmetric since uniform acceleration acts on it, thus time taken by bag to reach maximum height will be

  $t=\frac{2.2}{2}=1.1\text{s}$

Considering the upward motion of the bag, acceleration will be $a=-9.8{\text{m/s}}^{\text{2}}$ (negative sign denotes downward direction). Using the above-mentioned formula for the velocity, the initial velocity of the bag can be calculated as

  $\begin{array}{c}{v}_f=v_i+a\Delta t\\ 0=v_i+\left(-9.8\right)\left(1.1\right)\\ v_i=10.78\text{m/s}\end{array}$

Conclusion:

Thus, the initial velocity with which the beanbag was thrown is $10.78\text{m/s}$ .