Chapter 3, Problem 37P

(a)

To determine

To find: The horizontal distance travelled by the shot for $\theta =0^{\circ }$ .

Answer to Problem 37P

  $\text{x=5}\text{.00}\text{m}$

Explanation of Solution

Given data:

  $\begin{array}{c}\text{Initial}\text{velocity}\left(u\right)=8.00{\text{ms}}^{\text{-1}}\\ \text{Height}=1.90\text{m}\\ \text{Angle =}{\text{0}}^{\circ }\end{array}$

Formula used:

  $h=\frac{1}{2}g.t^2$

  $x=u.t$

Calculation:

  $\begin{array}{c}h=\frac{1}{2}g.t^2\\ 1.90m=\frac{1}{2}\times 9.81{\text{ms}}^{\text{-2}}.t^2\\ t^2=\frac{1.90m\times 2}{9.81{\text{ms}}^{\text{-2}}}\\ t^2=0.38{\text{s}}^{\text{2}}\\ t=0.622\text{s}\end{array}$

  $\begin{array}{l}x=u.t\\ x=8.00{\text{ms}}^{\text{-1}}\times 0.622\text{s}\\ \text{x=5}\text{.00}\text{m}\end{array}$

Conclusion:The horizontal distance travelled by the shot for $\theta =0^{\circ }$ is,

  $\text{x=5}\text{.00}\text{m}$

(b)

To determine

To find: The horizontal distance traveled by the shot for $\theta ={40}^{\circ }$ .

Answer to Problem 37P

  $\text{X=9}\text{.85 m}$

Explanation of Solution

Given data:

  $\begin{array}{c}\text{Initial}\text{velocity}\left(u\right)=8.00{\text{ms}}^{\text{-1}}\\ \text{Height}=1.90\text{m}\\ \text{Angle =}4{\text{0}}^{\circ }\end{array}$

Formula used:

  $V_x=V.\cos \theta$

  $V_y=V.\sin {40}^{\circ }$

  $V_{Final}=V_{Initial}-g.t$

  $h=V_{y}t+\frac{1}{2}g.t^2$

  $h=\frac{1}{2}g.t^2$

  $X=V_{x}t$

Calculation:

  

  $\begin{array}{l}{V}_x=V.\cos \theta \\ V_x=8\text{m}/s.\cos {40}^{\circ }\\ V_x=6.12{\text{ms}}^{\text{-1}}\end{array}$

  $\begin{array}{l}{V}_y=V.\sin {40}^{\circ }\\ V_y=8\text{m}/s.\sin {40}^{\circ }\\ V_y=5.14{\text{ms}}^{\text{-1}}\end{array}$

  $\begin{array}{c}{V}_{Final}=V_{Initial}-g.t\\ 0=5.14{\text{ms}}^{\text{-1}}-9.81{\text{ms}}^{\text{-2}}.t\\ t=0.52\text{s}\end{array}$

  $\begin{array}{c}h=V_{y}t+\frac{1}{2}g.t^2\\ h=\left(5.14\times 0.52\right)-\frac{1}{2}\mathrm{.9.81.}{\left(0.52s\right)}^2\\ h=1.35\text{m}\end{array}$

  $\begin{array}{l}{h}_{\max }=1.35\text{m}+1.9\text{m}\\ h_{\max }=3.25\text{m}\end{array}$

Free fall from the maximum height.

  $\begin{array}{c}h=\frac{1}{2}g.t^2\\ 3.25\text{m}=\frac{1}{2}\times 9.81\times t^2\\ t^2=.6625\\ t=\sqrt{.6625}\\ t=0.814\text{s}\end{array}$

  $\begin{array}{l}{t}_{Total}=0.814\text{s}+0.52\text{s}\\ t_{Total}=1.334\text{s}\end{array}$

  $\begin{array}{c}X=V_{x}t\\ X=6.12{\text{ms}}^{\text{-1}}\times 1.334\text{s}\\ \text{X=8}\text{.16 m}\end{array}$

Conclusion:

The horizontal distance traveled by the shot for $\theta ={40}^{\circ }$ .

  $\text{X}\text{=}8.16\text{ m}$

(c)

To determine

To find: The horizontal distance traveled by the shot for $\theta ={45}^{\circ }$ degrees.

Answer to Problem 37P

  $\text{X=9}\text{.83 m}$

Explanation of Solution

Given data:

  $\begin{array}{c}\text{Initial}\text{velocity}\left(u\right)=8.00{\text{ms}}^{\text{-1}}\\ \text{Height}=1.90\text{m}\\ \text{Angle =}4{\text{5}}^{\circ }\end{array}$

Formula used:

  $V_x=V.\cos \theta$

  $V_y=V.\sin {45}^{\circ }$

  $V_{Final}=V_{Initial}-g.t$

  $h=V_{y}t+\frac{1}{2}g.t^2$

  $h=\frac{1}{2}g.t^2$

  $X=V_{x}t$

Calculation:

  

  $\begin{array}{l}{V}_x=V.\cos \theta \\ V_x=8\text{m}/s.\cos {45}^{\circ }\\ V_x=5.65{\text{ms}}^{\text{-1}}\end{array}$

  $\begin{array}{l}{V}_y=V.\sin {45}^{\circ }\\ V_y=8\text{m}/s.\sin {45}^{\circ }\\ V_y=5.65{\text{ms}}^{\text{-1}}\end{array}$

  $\begin{array}{c}{V}_{Final}=V_{Initial}-g.t\\ 0=5.65{\text{ms}}^{\text{-1}}-9.81{\text{ms}}^{\text{-2}}.t\\ t=0.57\text{s}\end{array}$

  $\begin{array}{c}h=V_{y}t+\frac{1}{2}g.t^2\\ h=\left(5.65\times 0.57\right)-\frac{1}{2}\mathrm{.9.81.}{\left(0.57s\right)}^2\\ h=1.63\text{m}\end{array}$

  $\begin{array}{l}{h}_{\max }=1.63\text{m}+1.9\text{m}\\ h_{\max }=3.53\text{m}\end{array}$

Free fall from the maximum height.

  $\begin{array}{c}h=\frac{1}{2}g.t^2\\ 3.53\text{m}=\frac{1}{2}\times 9.81\times t^2\\ \\ t=.848\text{s}\end{array}$

  $\begin{array}{l}{t}_{Total}=.848\text{s}+0.57\text{s}\\ t_{Total}=1.42\text{s}\end{array}$

  $\begin{array}{c}X=V_{x}t\\ X=5.65{\text{ms}}^{\text{-1}}\times 1.42\text{s}\\ \text{X=8 m}\end{array}$

Conclusion:

The horizontal distance traveled by the shot for $\theta ={45}^{\circ }$ degrees is,

  $\text{X=8}\text{.0m}$