Question

(II) A 145-g baseball is dropped from a tree 12.0 m above the ground. (a) With what speed would it hit the ground if air resistance could be ignored? (b) If it actually hits the ground with a speed of 8.00 m/s, what is the average force of air resistance exerted on it?

Short answer

(a) The speed of the baseball is \(15.33\;\frac{{\rm{m}}}{{\rm{s}}}\)if air resistance could be ignored.

(b) The average air resistance force is 1.03 N (upward).

Step-by-step solution

Given data

The work done by the air resistive force is equal to the loss in mechanical energy.

Given data:

The mass of the baseball is given below:

\(\begin{aligned}m &= 145\;{\rm{g}}\\ &= 0.145\;{\rm{kg}}\end{aligned}\)

The initial height of the baseball is\(h = 12.0\,{\rm{m}}\).

The initial speed of the baseball is\({v_1} = 0\)as it starts from rest.

In the presence of air, the speed of the baseball is\({v_3} = 8.00\;\frac{{\rm{m}}}{{\rm{s}}}\)when it hits the ground.

Assumptions:

Let the speed of the baseball be\({v_2}\)when air resistance is ignored.

Let F be the average air resistance exerted on the baseball.

Calculation for part (a)

The total mechanical energy at the starting point is calculated as follows:

\(\begin{aligned}{E_1} &= mgh + \frac{1}{2}mv_1^2\\ &= mgh + \left( {\frac{1}{2}m \times {0^2}} \right)\\ &= mgh\end{aligned}\)

The mechanical energy of the baseball when it hits the ground is calculated as follows:

\(\begin{aligned}{E_2} &= \left( {mg \times 0} \right) + \frac{1}{2}mv_2^2\\ &= \frac{1}{2}mv_2^2\end{aligned}\)

From energy conservation,

\(\begin{aligned}{E_2} &= {E_1}\\\frac{1}{2}mv_2^2 &= mgh\\{v_2} &= \sqrt {2gh} \end{aligned}\)

Now substituting the values in the above equation,

\(\begin{aligned}{v_2} &= \sqrt {2 \times \left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right) \times \left( {12.0\,{\rm{m}}} \right)} \\ &= 15.33\;\frac{{\rm{m}}}{{\rm{s}}}\end{aligned}\)

Hence, the speed of the baseball is \(15.33\;\frac{{\rm{m}}}{{\rm{s}}}\)if air resistance could be ignored.

Calculation for part (b)

The total mechanical energy of the baseball when it hit the ground in the presence of air is calculated as follows:

\(\begin{aligned}{E_3} &= \left( {mg \times 0} \right) + \frac{1}{2}mv_3^2\\ &= \frac{1}{2}mv_3^2\end{aligned}\)

The work done by the resistive force of is \(W = Fh\) .

From energy conservation,

\(\begin{aligned}W &= {E_1} - {E_3}\\Fh &= mgh - \frac{1}{2}mv_3^2\\F &= mg - \frac{{mv_3^2}}{{2h}}\end{aligned}\)

Substituting the values in the above equation,

\(\begin{aligned}F &= \left[ {\left( {0.145\;{\rm{kg}}} \right) \times \left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right)} \right] - \left[ {\frac{{\left( {0.145\;{\rm{kg}}} \right){{\left( {8.00\;\frac{{\rm{m}}}{{\rm{s}}}} \right)}^2}}}{{2 \times 12.0\,{\rm{m}}}}} \right]\\ &= 1.03\;{\rm{N}}\end{aligned}\)

The resistance force is always against the direction of motion.

Hence, the average air resistance force is 1.03 N (upward).