Chapter 3, Problem 67GP

To determine

To calculate:The initial velocity and direction of the projectile.

Answer to Problem 67GP

The projectile is launched with initial velocity $63.05\text{ m/s}$ at an angle ${66}^{\text{o}}$ with the horizontal.

Explanation of Solution

Given data:

Distance between the projectile and the cliff, $\Delta x=195\text{ m}$

Height of the cliff, $\Delta y=155\text{ m}$

Flight time, $t=7.6\text{ s}$

Formula used:

The horizontal distance $\Delta x$ covered by a projectile during its motion is given by:

  $\Delta x=v_{x}t$

Where, $v_x$ is the velocity of the projectile in horizontal direction and t is the flight-time.

The vertical distance $\Delta y$ covered by a projectile during its motion is given by:

  $\Delta y=v_{y}t+\frac{1}{2}{a}_y{t}^2$

Where, $v_y$ is the velocity of the projectile in vertical direction, t is the flight-time, and $a_y$ is its acceleration in vertical direction.

Initial velocity of projectile, is $v_o=\sqrt{v_x{}^2+v_y{}^2}$ , where, $v_x$ and $v_y$ is the velocity of the projectile in horizontal and vertical direction respectively.

And the direction of projectile is given by:

  $\theta ={\tan }^{-1}\frac{v_y}{v_x}$

Calculation:

The projectile is at ground level.

Consider upward direction to be positive y direction.

Horizontal range of the projectile is the distance between the projectile and the cliff,that is,

  $\Delta x=195\text{ m}$

Flight time, $t=7.6\text{ s}$

Then, using horizontal range formula, initial velocity along horizontal direction is given by:

  $\begin{array}{l}\Delta x=v_{x}t\\ v_x=\frac{\Delta x}{t}=\frac{195\text{ m}}{7.6\text{ s}}=25.65\text{ m/s}\end{array}$

The vertical displacement of the projectile is equal to the height of the cliff, $\Delta y=155\text{ m}$

The acceleration in vertical direction, $a_y=-g$ ( g is the acceleration due to gravity; $g=9.8{\text{ m/s}}^2$ )

Using vertical distance formula, initial velocity along vertical direction is given by:

  $\begin{array}{l}\Delta y=v_{y}t+\frac{1}{2}{a}_y{t}^2\\ v_y=\frac{\Delta y-\frac{1}{2}{a}_y{t}^2}{t}\\ v_y=\frac{155\text{ m}-\frac{1}{2}\cdot (-9.8{\text{ m/s}}^2){(7.6\text{ s)}}^2}{7.6\text{ s}}\\ v_y=57.6\text{ m/s}\end{array}$

Therefore, initial velocity is:

  $\begin{array}{l}{v}_o=\sqrt{v_x{}^2+v_y{}^2}\\ v_o=\sqrt{{(25.65\text{ m/s})}^2+{(57.6\text{ m/s)}}^2}\\ v_o=63.05\text{ m/s}\end{array}$

And the direction of projectile is

  $\begin{array}{c}\theta ={\tan }^{-1}\frac{v_y}{v_x}\\ ={\tan }^{-1}\frac{57.6}{25.65}\\ ={\tan }^{-1}(2.245)={66}^{\text{o}}\end{array}$

Conclusion:

The projectile is launched with initial velocity $63.05\text{ m/s}$ at an angle ${66}^{\text{o}}$ with the horizontal.