Q16E - University Physics with Modern Physics

Question

On level ground a shell is fired with an initial velocity of $\mathbf{40}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$ at $\mathbf{60}\mathbf{°}$ above the horizontal and feels no appreciable air resistance. (a) Find the horizontal and vertical components of the shell’s initial velocity. (b) How long does it take the shell to reach its highest point? (c) Find its maximum height above the ground. (d) How far from its firing point does the shell land? (e) At its highest point, find the horizontal and vertical components of its acceleration and velocity

Step-by-Step Solution

Answer

a) The vertical and horizontal of the velocity of the shell is 34.6 m/s and 20 m/s  respectively.

b) The time taken by the shell to reach highest point is 3.53 s .

c) The maximum height of the shell above the ground is 61 m .

d) The distance of the shell landed from its firing point is 141 m .

e) The final horizontal and vertical component of acceleration are zero and $-9.8 \mathrm{m}/{\mathrm{s}}^2$, and the final vertical and horizontal component of velocity are zero and 20 m/s respectively.

Step 1: Identification of given data

The given data can be listed below,

• The initial velocity of the shell is, $v_0=40 \mathrm{m}/\mathrm{s}$.
• The horizontal angle is $60°$ . 

Step 2: Concept/Significance of the maximum height.

A maximum vertical component of the initial velocity results in the greatest height. The projectile is propelled vertically causes the phenomena.

Step 3: (a) Determination of the horizontal and vertical components of the shell’s initial velocity

The vertical component of velocity is given by,

${\mathrm{v}}_{0\mathrm{y}}={\mathrm{v}}_0\sin 60°$ 

Here, ${\mathrm{v}}_0$ is the initial velocity of shell.

Substitute value in the above,

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{y}}=\left(40 \mathrm{m}/\mathrm{s}\right)\sin 60° \\ =40 \mathrm{m}/\mathrm{s}\left(\frac{\sqrt{3}}{2}\right) \\ =34.6 \mathrm{m}.\mathrm{s} \end{gathered}$$ 

The horizontal component of velocity is given by,

 ${\mathrm{v}}_{0\mathrm{x}}={\mathrm{v}}_0\cos 60°$

 

Substitute all the values in the above,

 $$\begin{gathered} v_{0x}=\left(40 \mathrm{m}/\mathrm{s}\right)\cos 60° \\ =40 \mathrm{m}/\mathrm{s}\left(\frac{1}{2}\right) \\ =20 \mathrm{m}/\mathrm{s} \end{gathered}$$

Here, the vertical and horizontal of the velocity of the shell is 34.6 m/s and 20 m/s respectively.

Step 4: (b) Determination of the time taken by the shell to reach the highest point.

The time taken by the shell to reach highest point is given by,

$\mathrm{t}=\frac{{\mathrm{v}}_{\mathrm{y}}-{\mathrm{v}}_{0\mathrm{y}}}{{\mathrm{a}}_{\mathrm{y}}}$ 

Here, $v_y$ is the velocity in at maximum height, $v_{0y}$ is the vertical component of velocity and $a_y$ is the vertical component of acceleration.

 

Substitute all the values in the above,

$$\begin{gathered} \mathrm{t}=\frac{0-34.6 \mathrm{m}/\mathrm{s}}{-9.8 \mathrm{m}/{\mathrm{s}}^2} \\ =3.53 \mathrm{s} \end{gathered}$$ 

Thus, the time taken by the shell to reach highest point is 3.53 s .

Step 5: (c) Determination of the maximum height above the ground

The maximum height of the shell above the ground is given by,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{y}}^2={\mathrm{v}}_{0\mathrm{y}}^2+2{\mathrm{a}}_{\mathrm{y}}\mathrm{h} \\ {\mathrm{h}}_{\max }=\frac{-{\mathrm{v}}_{0\mathrm{y}}^2}{-2\mathrm{g}} \end{gathered}$$ 

Here, $v_y$ is the velocity in at maximum height, $v_{0y}$ is the vertical component of velocity and $a_y$ is the vertical component of acceleration denoted by g.

 

Substitute all the values in the above,

$$\begin{gathered} {\mathrm{h}}_{\max }=\frac{-{\left(34.6 \mathrm{m}/\mathrm{s}\right)}^2}{-2\times 9.8 \mathrm{m}/{\mathrm{s}}^2} \\ =61\mathrm{m} \end{gathered}$$  

Thus, the maximum height of the shell above the ground is 61 m .

Step 6: (d) Determination of the distance shell from its firing point:

The distance of shell from its firing point is given by,

$\mathrm{d}=2{\mathrm{v}}_{0\mathrm{x}}\mathrm{t}$ 

Here, $v_{0x}$ is the horizontal component of shell’s velocity and t is the time taken to reach the peak point.

 

Substitute all the values in the above,

$$\begin{gathered} \mathrm{d}=20 \mathrm{m}/\mathrm{s}\left(2\times 3.53 \mathrm{s}\right) \\ =141 \mathrm{m} \end{gathered}$$ 

Thus, the distance of the shell landed from its firing point is 141 m .

Step 7: (e) Determination of the horizontal and vertical components of its acceleration and velocity:

The horizontal acceleration of the shell is zero because there is no change in the horizontal velocity. The vertical component of the acceleration is the acceleration due to gravity acting on the shell which is given by,

$$\begin{gathered} {\mathrm{a}}_{\mathrm{y}}=-\mathrm{g} \\ =-9.8 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

At the maximum height the final vertical and horizontal velocity of shell are given by,

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{y}}=0 \\ {\mathrm{v}}_{\mathrm{x}}={\mathrm{v}}_{0\mathrm{y}} \\ =20 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Thus, the final horizontal and vertical component of acceleration are zero and$-9.8 \mathrm{m}/{\mathrm{s}}^2$ and the final vertical and horizontal component of velocity are zero and 20 m/s respectively.