Q69P - University Physics with Modern Physics

Question

In the middle of the night you are standing a horizontal distance of 14.0m from the high fence that surrounds the estate of your rich uncle. The top of the fence is 5.00m above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of 1.60m above the ground and at an angle of 56.$\mathbf{0}\mathbf{°}$above the horizontal. (a) What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? (b) For the initial velocity calculated in part (a), what horizontal distance beyond the fence will the rock land on the ground?

Step-by-Step Solution

Answer

(a) u = 13.3 m/s

 

(b) 3.8m

Step 1: Newton’s law

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ The velocity contains two components, one is a horizontal component and another one is vertical. 

 

According to newton’s laws of motion,

 

 

Where, s, u, t, a, are displacement, initial velocity, time, and acceleration.

 

For the vertical and horizontal motion, the velocity component will be respectively.

Step 2: Given

The horizontal distance of self from the fence = 14.0m

 

The top of the fence = 5.00m

 

Height from which the rock threw = 1.60m

 

Angle = $56°$

Step 3: (a) Minimum initial speed

The vertical displacement would be 

                                                                              

$$\begin{gathered} {\mathrm{s}}_{\mathrm{v}}=5.00\mathrm{m}-1.60\mathrm{m} \\ {\mathrm{s}}_{\mathrm{v}}=3.40\mathrm{m} \end{gathered}$$

 

The horizontal Motion, gravity will be zero here

$$\begin{gathered} {\mathrm{s}}_{\mathrm{h}}=\mathrm{u} \mathrm{cos\theta t} \\ 14\mathrm{m}=\mathrm{ucos}56°\times \mathrm{t} \\ \mathrm{t}=\frac{25.04}{\mathrm{u}}\mathrm{s} \end{gathered}$$ 

The vertical motion,

           

                 $$\begin{gathered} {\mathrm{s}}_{\mathrm{v}}=\mathrm{usin\theta t}+\frac{1}{2}{\mathrm{at}}^2 \\ 3.40\mathrm{m}=\mathrm{u} \sin 56°\times \frac{25.04}{\mathrm{u}}\times -9.8\mathrm{m}/{\mathrm{s}}^2\times {\left(\frac{25.04}{\mathrm{u}}\right)}^2 \\ \mathrm{u}=13.3\mathrm{m}/\mathrm{s} \end{gathered}$$

         

Step 4: (b) Horizontal distance beyond the fence traveled by rock

From the vertical motion,                                                                

$$\begin{gathered} \mathrm{s}=\mathrm{u} \mathrm{sin\theta t}+\frac{1}{2}{\mathrm{at}}^2 \\ -1.60\mathrm{m}=13.3\mathrm{m}/\mathrm{s}\times \sin 56°\times \mathrm{t}+\frac{1}{2}\times -9.8\mathrm{m}/{\mathrm{s}}^2\times \mathrm{t} \\ \mathrm{t}=2.388\mathrm{s} \end{gathered}$$

 

From the horizontal motion 5,

$$\begin{gathered} \mathrm{s}=\mathrm{u} \cos \mathrm{\theta t}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{s}=13. \mathrm{m}/\mathrm{s}\times \cos 56°\times \mathrm{t}+\frac{1}{2}\times 2.38\mathrm{s} \\ \mathrm{t}=17.8\mathrm{m} \end{gathered}$$

This the instant horizontal position from the launch point, so the horizontal distance beyond the fence will be,

 17.8m - 14.0m = 3.8m