Question

A salesman is standing on the Golden Gate Bridge in a traffic jam. He is at a height of 71.8 m above the water below. He receives a call on his cell phone that makes him so mad that he throws his phone horizontally off the bridge with a speed of 23.0 m/s. (a) How far does the cell phone travel horizontally before hitting the water? (b) What is the speed with which the phone hits the water?

Answer 1

To solve how far the cell phone travels horizontally before hitting the water, we calculate the free-fall time for 71.8 meters height and then use it to find the range. For the impact speed, we find the vertical speed at impact and use the Pythagorean theorem with the constant horizontal speed to find the resultant impact speed.

Step-by-step explanation:

(a) The horizontal distance (\(d\)) traveled by the cell phone can be calculated using the horizontal motion equation:

`\[ d = v_0 \cdot t \]`

where \(v_0\) is the initial horizontal velocity and \(t\) is the time of flight. In this case, the initial horizontal velocity is the given speed of the phone (23.0 m/s), and the time of flight (\(t\)) can be found using the vertical motion equation:

`\[ h = (1)/(2) g t^2 \]`

where \(h\) is the height of the bridge, \(g\) is the acceleration due to gravity (approximately 9.8 m/s²). Solving for \(t\), we find \(t = \sqrt{\frac{2h}{g}}\). Substituting this into the first equation, we get:

`\[ d = v_0 \cdot \sqrt{(2h)/(g)} \]`

Substituting the known values, we can calculate \(d\).

(b) The final velocity (\(v_f\)) just before hitting the water can be found using the vertical motion equation:

`\[ v_f = √(v_0^2 + 2gh) \]`

Substituting the known values, we can calculate \(v_f\)