Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Expert Solution & Answer
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 ( vo )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:
y=yo+voyt+12gt2y=yo+vosinθ t+12gt2 ... (1)
On the other hand, the horizontal distance ( x ) is defined as:
x=voxtx=vocosθ t ...(2)
Where:
- y is vertical position reached by the diver.
- yo is in the initial vertical position from where the diver was launched.
- voy is the initial vertical speed of the diver.
- g is acceleration due to gravity on Earth.
- t is the time elapsed.
- vo is theinitial speed of the diver.
- θ is the diver’s launch angle.
- x is the horizontal distance.
- vox is the initial horizontal speed of the diver.
Calculations: Isolate yo from equation (1):
yo=y−voysinθt+12gt2 ...(3)
Substitute values in equation (3)
yo=0−2.5sin0o(3)+12(9.8)(3.0)2 ...(5)
yo=44 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=2.5cos0o(3.0)=7.5 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.
Chapter 3 Solutions
Physics: Principles with Applications
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