Chapter 2, Problem 47P

(a)

To determine

The time at which the stone reach the bottom of the cliff.

Answer to Problem 47P

The time at which the stone reach the bottom of the cliff is 5.2 s.

Explanation of Solution

Given:

The initial velocity is $v_0=12.0\text{m}/\text{s}$ .

The height of the cliff is $y=70\text{ m}$ .

Formula used:

The expression for the height is,

  $y=y_0+v_{0}t+\frac{1}{2}g{t}^2$

Here, $y_0$ is the initial height, $v_0$ is the initial velocity, t is the time, and g is the acceleration due to gravity.

Calculation:

Consider the initial height of the ball $y_0=0$ .

Consider the acceleration due to gravity is $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

The time at which the stone reach the bottom of the cliff using the relation is,

  $\begin{array}{l}y=y_0+v_{0}t+\frac{1}{2}g{t}^2\\ -70\text{ m}=0+\left(12\text{ m/s}\right)t+\frac{1}{2}\left(-9.8{\text{ m/s}}^2\right)t^2\\ 4.9t^2-12t-70=0\\ t=5.2\text{ s}\end{array}$

Conclusion:

Thus, the time at which the stone reach the bottom of the cliff is 5.2 s.

(b)

To determine

The speed of the ball just before hitting.

Answer to Problem 47P

The speed of the ball just before hitting is 38.93 m/s.

Explanation of Solution

Given:

The initial velocity is $v_0=12.0\text{m}/\text{s}$ .

The height of the cliff is $y=70\text{ m}$ .

Formula used:

The expression for the velocity is,

  $v^2=v_0^2+2as$

Here, $v_0$ is the initial velocity, a is the acceleration due to gravity, and s is the height of the ball from the cliff to top.

The expression for velocity is,

  $v=\sqrt{2gh}$

Here, g is the acceleration due to gravity and h is the height.

Calculation:

Consider the velocity $v=0$ at a maximum height.

The height of the ball above the cliff can be calculated as,

  $\begin{array}{l}\left(0\right)={\left(12\text{ m/s}\right)}^2+2\times \left(-9.8{\text{ m/s}}^2\right)s\\ s=7.35\text{ m}\end{array}$

The total height the ball reaches will be,

  $\begin{array}{l}h=y+s\\ h=\left(70\text{ m}\right)+\left(7.35\text{ m}\right)\\ h=77.35\text{ m}\end{array}$

The speed of the ball just before hitting is,

  $\begin{array}{c}v=\sqrt{2gh}\\ =\sqrt{2\times \left(9.8{\text{ m/s}}^2\right)\times \left(77.35\text{ m}\right)}\\ =38.93\text{ m/s}\end{array}$

Conclusion:

Thus, the speed of the ball just before hitting is 38.93 m/s.

(c)

To determine

The total distance travelled by the ball.

Answer to Problem 47P

The total distance travelled by the ball is 77.35 m.

Explanation of Solution

Given:

The initial velocity is $v_0=12.0\text{m}/\text{s}$ .

The height of the cliff is $y=70\text{ m}$ .

Formula used:

The expression for the velocity is,

  $v^2=v_0^2+2as$

Here, $v_0$ is the initial velocity, a is the acceleration due to gravity, and s is the height of the ball from the cliff to top.

Calculation:

Refer to part (b).

The total height the ball reaches will be,

  $\begin{array}{l}h=y+s\\ h=\left(70\text{ m}\right)+\left(7.35\text{ m}\right)\\ h=77.35\text{ m}\end{array}$

So, the total distance traveled by the ball is 77.35 m.

Conclusion:

Thus, the total distance traveled by the ball is 77.35 m.