Chapter 3, Problem 50P

(a)

To determine

The time taken by the rock thrown from the edge of a cliff to reach the ground.

Answer to Problem 50P

The rock takes 2.93 s to reach the ground.

Explanation of Solution

Given:

The depth of the ground from the top of the cliff, $y=-20.0\text{ m}$

Velocity of projection

, $v_0=15.0\text{ m/s}$

Angle of projection, ${\theta }_0=30.0°$

Formula used:

The time taken to reach the ground is determined using the expression

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

Here, $v_{0y}$ is the vertical component of the rock’s velocity, given by

  $v_{0y}=v_0\cos {\theta }_0$.....(2)

Calculation:

The path of the rock is shown in the diagram below:

  

Calculate the value of $v_{0y}$ by substituting the given values in equation (2).

  $\begin{array}{c}{v}_{0y}=v_0\sin {\theta }_0\\ =\left(15.0\text{ m/s}\right)\left(\sin 30.0°\right)\\ =7.5\text{ m/s}\end{array}$

Substitute the values of variables in equation (1).

  $\begin{array}{l}y=v_{0y}t-\frac{1}{2}g{t}^2\\ \left(-20.0\text{ m}\right)=\left(7.5\text{ m/s}\right)t-\frac{1}{2}\left(9.8{\text{ m/s}}^2\right)t^2\end{array}$

Rewrite the equation as follows:

  $\left(4.9{\text{ m/s}}^2\right)t^2-\left(7.5\text{ m/s}\right)t-\left(20.0\text{ m}\right)=0$

Solve the quadratic equation.

  $\begin{array}{c}t=\frac{-\left(-7.5\text{ m/s}\right)\pm \sqrt{{\left(-7.5\text{ m/s}\right)}^2-4\left(4.9{\text{ m/s}}^2\right)\left(-20.0\text{ m}\right)}}{2\left(4.9{\text{ m/s}}^2\right)}\\ =\frac{\left(7.5\text{ m/s}\right)\pm \left(21.17\text{ m/s}\right)}{\left(9.8{\text{ m/s}}^2\right)}\end{array}$

Take the positive root, since time cannot have a negative value.

  $t=2.93\text{ s}$

Conclusion:

Thus, the rock takes 2.93 s to reach the ground.

(b)

To determine

The horizontal distance between the point of projection and the point where the rock lands on the ground.

Answer to Problem 50P

The rock lands at a horizontal distance of 38.1 m from the point of projection.

Explanation of Solution

Given:

Velocity of projection, $v_0=15.0\text{ m/s}$

Angle of projection, $\theta =30.0°$

Time of flight of the rock, $t=2.93\text{ s}$

Formula used:

The horizontal component of the rock’s velocity is constant, since no force acts on the rock in the horizontal direction in the absence of air resistance.

The horizontal distance travelled by the rock in a time t is given by,

  $x=v_{0x}t$.....(3)

Here, the horizontal component $v_{0x}$ is given by

  $v_{0x}=v_0\cos {\theta }_0$.....(4)

Calculation:

Substitute the given values in equation (4) and calculate the value of $v_{0x}$ .

  $\begin{array}{c}{v}_{0x}=v_0\cos {\theta }_0\\ =\left(15.0\text{ m/s}\right)\left(\cos 30.0°\right)\\ =13.0\text{ m/s}\end{array}$

Substitute the values of $v_{0x}$ and t in equation (3) and calculate the value of x .

  $\begin{array}{c}x=v_{0x}t\\ =\left(13.0\text{ m/s}\right)\left(2.93\text{ s}\right)\\ =38.1\text{ m}\end{array}$

Conclusion:

Thus, the rock lands at a horizontal distance of 38.1 m from the point of projection.

(c)

To determine

The velocity of the rock when it lands on the ground.

Answer to Problem 50P

The velocity of the rock when it lands on the ground is found to have a magnitude $-21.2\text{ m/s}$ at an angle $-58.5°$ to the horizontal.

Explanation of Solution

Given:

The horizontal component of the rock’s initial velocity, $v_{0x}=13.0\text{ m/s}$

The vertical component of the rock’s initial velocity, $v_{0y}=7.5\text{ m/s}$

Time of flight of the rock, $t=2.93\text{ s}$

Formula used:

The vertical component of the rock’s velocity when it reaches the ground is given by the expression,

  $v_y=v_{0y}-gt$.....(5)

The horizontal component of the rock’s velocity when it reaches the ground remains unchanged during its motion. Therefore,

  $v_x=v_{0x}$.....(6)

The magnitude of the velocity of the particle when it reaches the ground is given by,

  $v=\sqrt{v_x^2+v_y^2}$.....(7)

The angle made by the velocity vector to the horizontal is given by the expression,

  $\alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$.....(8)

Calculation:

Calculate the vertical component of the rock’s velocity when it reaches the ground by substituting the values of the variables in equation (5).

  $\begin{array}{c}{v}_y=v_{0y}-gt\\ =\left(7.5\text{ m/s}\right)-\left(9.8{\text{ m/s}}^2\right)\left(2.93\text{ s}\right)\\ =-21.2\text{ m/s}\end{array}$

Since $v_x=v_{0x}$ , therefore,

  $v_x=13.0\text{ m/s}$

Substitute the values of $v_x$ and $v_y$ in equation (7) and calculate the magnitude of v .

  $\begin{array}{c}v=\sqrt{v_x^2+v_y^2}\\ =\sqrt{{\left(13.0\text{ m/s}\right)}^2+{\left(-21.2\text{ m/s}\right)}^2}\\ =24.7\text{ m/s}\end{array}$

Substitute the values of $v_x$ and $v_y$ in equation (8) and calculate the value of $\alpha$ .

  $\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)\\ ={\tan }^{-1}\left(\frac{-21.2\text{ m/s}}{13.0\text{ m/s}}\right)\\ =-58.5°\end{array}$

Conclusion:

Thus the velocity of the rock when it lands on the ground is found to have a magnitude $-21.2\text{ m/s}$ at an angle $-58.5°$ to the horizontal.