Chapter 3, Problem 19P

Solution

To Determine:Height of the cliff and the distance from its base where the diver hit the water.

Solution:The height of the cliff is 44 m and the diver hit the water at a distance of 7.5 m from the base of the cliff.

Explanation:

Given:

The initial speed ( $v_o$ )of the diver is 2.5 m/s.The diver is thrown horizontally and hits water intime tof 3.0 s.

Formula used:

From kinematic equation:

  $\begin{array}{l}y=y_o+v_{oy}t+\frac{1}{2}g{t}^2\\ y=y_o+v_o\sin \theta t+\frac{1}{2}g{t}^2\mathrm{...}\left(1\right)\end{array}$

On the other hand, the horizontal distance ( x ) is defined as:

  $\begin{array}{l}x=v_{ox}t\\ x=v_o\cos \theta t\mathrm{...}\left(2\right)\end{array}$

Where:

• y is vertical position reached by the diver.
• ${\text{y}}_o$ is in the initial vertical position from where the diver was launched.
• $v_{oy}$ is the initial vertical speed of the diver.
• g is acceleration due to gravity on Earth.
• t is the time elapsed.
• $v_o$ is theinitial speed of the diver.
• $\theta$ is the diver’s launch angle.
• x is the horizontal distance.
• $v_{ox}$ is the initial horizontal speed of the diver.

Calculations: Isolate ${\text{y}}_o$ from equation (1):

  $y_o=y-v_{oy}\sin \theta t+\frac{1}{2}g{t}^2\mathrm{...}\left(3\right)$

Substitute values in equation (3)

  $y_o=0-2.5\sin 0^o\left(3\right)+\frac{1}{2}\left(9.8\right){\left(3.0\right)}^2\mathrm{...}\left(5\right)$

  $y_o=44\text{m}$

Therefore, the height of the cliff is 44 m.

On the other hand, in the case of horizontal distance x , substitute values in equation (2):

  $x=2.5\cos 0^o\left(3.0\right)=7.5\text{m}$

Therefore, the diver hits the water at a distance of 7.5 m from the base of the cliff.

Conclusion:

Thus,the height of the cliff is 44 m and the diver hits the water at a distance of 7.5 m from the base of the cliff.