Chapter 3, Problem 96P

(a)

To determine

The vertical component of initial velocity.

Explanation of Solution

Given:

The time is $20.0\text{s}$.

The distance is $3000\text{m}$ above the horizontal.

The distance is $450\text{m}$ in vertical direction.

Formula used:

Write the expression for the vertical displacement.

  $\Delta y=v_{0y}\Delta t+\frac{1}{2}{a}_y{\left(\Delta t\right)}^2$

Here, $\Delta y$ is the displacement is y direction, $v_{0y}$ is the initial velocity in the y direction, $a_y$ is the acceleration in $y$ direction and $\Delta t$ is the time.

Substitute $-g$ for $a_y$ in the above equation.

  $\Delta y=v_{0y}\Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2$

Solve the above equation for $v_{0y}$ .

  $v_{0y}=\frac{\Delta y+\frac{1}{2}g{\left(\Delta t\right)}^2}{\Delta t}$ ....... (1)

Calculation:

Substitute $20.0\text{s}$ for $\Delta t$ , $450\text{m}$ for $\Delta y$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (1).

  $\begin{array}{l}{v}_{0y}=\frac{450\text{m}+\frac{1}{2}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right){\left(20.0\text{s}\right)}^2}{20.0\text{s}}\\ v_{0y}=120.6\text{m}/\text{s}\end{array}$

Conclusion:

The vertical component of the initial velocity is $120.6\text{m}/\text{s}$ .

(b)

To determine

The horizontal component of initial velocity.

Explanation of Solution

Given:

The time is $20.0\text{s}$.

The distance is $3000\text{m}$ above the horizontal.

The distance is $450\text{m}$ in vertical direction.

Formula used:

Write the expression for the velocity.

  $v=\frac{\Delta x}{\Delta t}$ ....... (2)

Here, $v$ is the velocity, $\Delta x$ is in the position in the x direction and $\Delta t$ is the time.

Calculation:

Substitute $3000\text{m}$ for $\Delta x$ and $20.0\text{s}$ for $\Delta t$ in equation (2).

  $\begin{array}{l}v=\frac{3000\text{m}}{20.0\text{s}}\\ v=150\text{m}/\text{s}\end{array}$

Conclusion:

The horizontal component of initial velocity is $150\text{m}/\text{s}$ .

(c)

To determine

The maximum height above the launch point.

Explanation of Solution

Given:

The time is $20.0\text{s}$.

The distance is $3000\text{m}$ above the horizontal.

The distance is $450\text{m}$ in vertical direction.

Formula used:

Write the expression for the vertical displacement.

  $h=v_{0y}\Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2$

Here, $h$ is the height.

Substitute $v_0\sin \theta$ for $v_{0y}$ in the above equation.

  $h=\left(v_0\sin \theta \right)\Delta t-\frac{1}{2}g{\left(\Delta t\right)}^2$ ....... (3)

Write the expression for the relation of initial and final velocity.

  $v_y=v_{0y}+a_y\Delta t$

Here, $v_{0y}$ is the initial velocity and $a_y$ is the acceleration.

Substitute $v_0\sin \theta$ for $v_{0y}$ in the above equation and $-g$ for $a_y$ in the above equation.

  $v_y=\left(v_0\sin \theta \right)-g\Delta t$ ....... (4)

Solve the above equation for $\Delta t$ .

  $\Delta t=\frac{v_0\sin {\theta }_0}{g}$

Substitute $\frac{v_0\sin {\theta }_0}{g}$ for $\Delta t$ in equation (3).

  $h=\left(v_0\sin \theta \right)\Delta t-\frac{1}{2}g{\left(\frac{v_0\sin {\theta }_0}{g}\right)}^2$

Simplify the above equation.

  $h=\frac{v_0{}^2{\sin }^2{\theta }_0}{2g}$ ....... (5)

Write the expression for the resultant velocity.

  $v=\sqrt{{\left(v_{0x}\right)}^2+{\left(v_{0y}\right)}^2}$ ....... (6)

Here, $v_{0x}$ is the component of velocity in the x direction and $v_{0y}$ is the component of velocity in the y direction.

Write the expression for the angle.

  $\theta ={\tan }^{-1}\left(\frac{v_{0y}}{v_{0x}}\right)$ ....... (7)

Write the expression for the horizontal displacement of the projectile.

  $\Delta x=v_{0x}\Delta t$

Here, $\Delta x$ is the displacement and $v_{0x}$ is the initial velocity in the x direction.

Substitute $v_0\cos {\theta }_0$ for $v_{0x}$ in the above equation.

  $\Delta x=v_0\cos {\theta }_0\Delta t$

Solve the above equation for $\theta$ .

  ${\theta }_0={\cos }^{-1}\left(\frac{\Delta x}{v_0\Delta t}\right)$ ....... (8)

Calculation:

Substitute $150\text{m}/\text{s}$ for $v_{0x}$ and $120.6\text{m}/\text{s}$ for $v_{0y}$ in equation (5).

  $\begin{array}{l}v=\sqrt{{\left(150\text{m}/\text{s}\right)}^2+{\left(120.6\text{m}/\text{s}\right)}^2}\\ v=192.5\text{m}/\text{s}\end{array}$

Substitute $3000\text{m}$ for $\Delta x$ , $20.0\text{s}$ for $\Delta t$ and $192.5\text{m}/\text{s}$ for $v$ in equation (6).

  $\begin{array}{l}\theta ={\cos }^{-1}\left(\frac{3000\text{m}}{192.5\text{m}/\text{s}\left(20.0\text{s}\right)}\right)\\ \theta =38.80°\end{array}$

Substitute $38.80°$ for ${\theta }_0$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $192.5\text{m}/\text{s}$ for $v_0$ in equation(5).

  $\begin{array}{l}h=\frac{{\left(192.5\text{m}/\text{s}\right)}^2{\sin }^2\left(38.80°\right)}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ h=742\text{m}\end{array}$

Conclusion:

The maximum height above the launch point is $742\text{m}$ .

(d)

To determine

The speed and the angle that the velocity makes with the vertical.

Explanation of Solution

Given:

The time is $20.0\text{s}$.

The distance is $3000\text{m}$ above the horizontal.

The distance is $450\text{m}$ in vertical direction.

Formula used:

Write the expression for the relation of initial and final velocity.

  $v_y=v_{0y}+a_y\Delta t$

Here, $v_{0y}$ is the initial velocity and $a_y$ is the acceleration.

Substitute $v_0\sin \theta$ for $v_{0y}$ in the above equation and $-g$ for $a_y$ in the above equation.

  $v_y=\left(v_0\sin \theta \right)-g\Delta t$

Calculation:

Substitute $38.80°$ for ${\theta }_0$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $20.0\text{s}$ for $\Delta t$ and $192.5\text{m}/\text{s}$ for $v_0$ in equation (4).

  $\begin{array}{l}{v}_y=192.5\text{m}/\text{s}\left(\sin \left(38.80°\right)\right)-\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)20.0\text{s}\\ v_y=-75.60\text{m}/\text{s}\end{array}$

The speed of the projectile at impact is:

Substitute $150\text{m}/\text{s}$ for $v_x$ and $-75.60\text{m}/\text{s}$ for $v_y$ in equation (6).

  $\begin{array}{l}v=\sqrt{{\left(150\text{m}/\text{s}\right)}^2+{\left(-75.60\text{m}/\text{s}\right)}^2}\\ v=168\text{m}/\text{s}\end{array}$

The angle for impact is:

Substitute $150\text{m}/\text{s}$ for $v_x$ and $-75.60\text{m}/\text{s}$ for $v_y$ in equation (7).

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{-75.60\text{m}/\text{s}}{150\text{m}/\text{s}}\right)\\ \theta =-26.75°\end{array}$

Conclusion:

The speed and the angle that velocity makes with the vertical is $168\text{m}/\text{s}$ and $-26.75°$ respectively.