Chapter 3, Problem 36P

(a)

To determine

To Find:The range of a ball thrown horizontally while standing on level ground.

Answer to Problem 36P

  $17.13\text{ m}$

Explanation of Solution

Given data:

The ball is thrown horizontally.

Formula used:

Second equation of motion:

  $s=ut+\frac{1}{2}a{t}^2$s

Calculation:

Applying $s=ut+\frac{1}{2}a{t}^2$ in the horizontal direction,

  $R=vt$ (there is no acceleration in the horizontal direction)

Applying $s=ut+\frac{1}{2}a{t}^2$ in the vertical downward direction

  $\begin{array}{l}h=\frac{1}{2}g{t}^2\\ t=\sqrt{\frac{2h}{g}}\end{array}$

Substituting value of t in $R=vt$

  $R=v\times \sqrt{\frac{2h}{g}}$

Assume that the person can throw the ball at 60 mph.

  $\begin{array}{c}v=60\text{mph }\\ \text{= 60}\times \text{0}\text{.44704}\text{m}{\text{s}}^{\text{-1}}\\ =26.82{\text{m s}}^{\text{-1}}\end{array}$

Assume that the person releases the ball from a height of 2 m.

  $h=2\text{ m}$

Substituting the values for v and h

  $\begin{array}{l}R=26.82\times \sqrt{\frac{2\times 2}{9.81}}\text{ m}\\ R=17.13\text{ m}\end{array}$

Conclusion:

Thus, the range of a ball thrown horizontally while standing on level ground = $17.13\text{ m}$ .

(b)

To determine

To Find:The range of a ball thrown at an angle 45° above the horizontal while standing on level ground

Answer to Problem 36P

  $74.10\text{ m}$

Explanation of Solution

Given data:

The ball is thrown at an angle 45° above the horizontal.

Formula used:

  $s=ut+\frac{1}{2}a{t}^2$s

Calculation:

Assume that the person can throw the ball at 60 mph.

  $\begin{array}{c}v=60\text{mph }\\ \text{= 60}\times \text{0}\text{.44704}\text{m}{\text{s}}^{\text{-1}}\\ =26.82{\text{m s}}^{\text{-1}}\end{array}$

Assume that the person releases the ball from a height of 2 m.

  $h=2\text{ m}$

Let the angle of release of the ball from the horizontalbe $\theta$.

Applying $s=ut+\frac{1}{2}a{t}^2$ in the horizontal direction,

  $R=v\cos \theta t$

Applying $s=ut+\frac{1}{2}a{t}^2$ in the vertical downward direction

  $h=-v\sin \theta t+\frac{1}{2}g{t}^2$

Substituting value of t in $R=v\cos \theta t$ to $h=-v\sin \theta t+\frac{1}{2}g{t}^2$

  $\begin{array}{l}h=-v\sin \theta \times \frac{R}{v\cos \theta }+\frac{1}{2}g{\left(\frac{R}{v\cos \theta }\right)}^2\\ 2h{v}^2{\cos }^2\theta =-v^{2}R\sin 2\theta +g{R}^2\end{array}$

Solving for R

  $R=\frac{v^2\sin 2\theta \pm \sqrt{v^4{\sin }^{2}2\theta +4g2h{v}^2{\cos }^2\theta }}{2g}$

There are 2 values for R. Taking the larger value,

  $R=\frac{v^2\sin 2\theta +\sqrt{v^4{\sin }^{2}2\theta +4g2h{v}^2{\cos }^2\theta }}{2g}$

Substituting values for v, h, $\theta$ and g

  $\begin{array}{l}R=\frac{{\left(26.82\right)}^2\sin \left(2\times {45}^{\circ }\right)+\sqrt{{\left(26.82\right)}^4{\sin }^2\left(2\times {45}^{\circ }\right)+8\times 9.81\times 2\times {\left(26.82\right)}^2{\cos }^2{45}^{\circ }}}{2\times 9.81}\text{ m}\\ R=74.10\text{ m}\end{array}$

Conclusion:

The range of a ball thrown at an angle 45° above the horizontal = $74.10\text{ m}$

(c)

To determine

The range of a ball thrown horizontallyfrom the top of a building $12m$ high

Answer to Problem 36P

The range of a ball thrown horizontally from the top of a building = $45.31\text{}\text{ m}$

Explanation of Solution

Given data:

The ball is thrown horizontallyfrom the top of a building $12m$ high.

Formula used:

  $s=ut+\frac{1}{2}a{t}^2$s

Calculation:

Assume that the person can throw the ball at 60 mph.

  $v=60\text{mph = 60}\times \text{0}\text{.44704}\text{m}{\text{.s}}^{\text{-1}}=26.82\text{m}{\text{.s}}^{\text{-1}}$

Assume that the person releases the ball from a height of 2 m.

  $h=2\text{ m}$

Applying $s=ut+\frac{1}{2}a{t}^2$ in the horizontal direction,

  $R=vt$

Applying $s=ut+\frac{1}{2}a{t}^2$ in the vertical downward direction

  $h+12=\frac{1}{2}g{t}^2$

Substituting value of t in $R=vt$ to $h+12=\frac{1}{2}g{t}^2$

  $\begin{array}{l}h+12=\frac{1}{2}g{\left(\frac{R}{v}\right)}^2\\ R=\sqrt{\frac{2v\left(h+12\right)}{g}}\end{array}$

Substituting values for v, h,and g

  $\begin{array}{l}R=\sqrt{\frac{2\times {\left(26.82\right)}^2\times \left(2+12\right)}{9.81}}\text{}\text{ m}\\ R=45.31\text{}\text{ m}\end{array}$

Conclusion:

The range of a ball thrown horizontally from the top of a building = $45.31\text{}\text{ m}$

(d)

To determine

The range of a ball thrown at an angle 45° above the horizontal from the top of a building 12 m high

Answer to Problem 36P

The range of a ball thrown at an angle 45° above the horizontal from the top of a building 12 m high = $85.35\text{ m}$

Explanation of Solution

Given data

The ball is thrown at an angle 45° above the horizontal from the top of a building 12 m high.

Formula used

  $s=ut+\frac{1}{2}a{t}^2$s

Calculation

Assume that the person can throw the ball at 60 mph.

  $v=60\text{mph = 60}\times \text{0}\text{.44704}\text{m}{\text{.s}}^{\text{-1}}=26.82\text{m}{\text{.s}}^{\text{-1}}$

Assume that the person releases the ball from a height of 2 m.

  $h=2\text{ m}$

Let the angle of release of the ball from the horizontalbe $\theta$.

Applying $s=ut+\frac{1}{2}a{t}^2$ in the horizontal direction,

  $R=v\cos \theta t$

Applying $s=ut+\frac{1}{2}a{t}^2$ in the vertical downward direction

  $h+12=-v\sin \theta t+\frac{1}{2}g{t}^2$

Substituting value of t in $R=v\cos \theta t$ to $h=-v\sin \theta t+\frac{1}{2}g{t}^2$

  $\begin{array}{l}h+12=-v\sin \theta \times \frac{R}{v\cos \theta }+\frac{1}{2}g{\left(\frac{R}{v\cos \theta }\right)}^2\\ 2\left(h+12\right)v^2{\cos }^2\theta =-v^{2}R\sin 2\theta +g{R}^2\end{array}$

Solving for R

  $R=\frac{v^2\sin 2\theta \pm \sqrt{v^4{\sin }^{2}2\theta +4g\times 2\left(h+12\right)v^2{\cos }^2\theta }}{2g}$

There are 2 values for R. Taking the larger value,

  $R=\frac{v^2\sin 2\theta +\sqrt{v^4{\sin }^{2}2\theta +4g\times 2\left(h+12\right)v^2{\cos }^2\theta }}{2g}$

Substituting values for v, h, $\theta$ and g

  $\begin{array}{l}R=\frac{{\left(26.82\right)}^2\sin \left(2\times {45}^{\circ }\right)+\sqrt{{\left(26.82\right)}^4{\sin }^2\left(2\times {45}^{\circ }\right)+8\times 9.81\times \left(2+12\right)\times {\left(26.82\right)}^2{\cos }^2{45}^{\circ }}}{2\times 9.81}\text{ m}\\ R=85.35\text{ m}\end{array}$

Conclusion

The range of a ball thrown at an angle 45° above the horizontal from the top of a building 12 m high = $85.35\text{ m}$