A person dives off the edge of a cliff 33m above the surface of the sea below. Assuming that air resistance is negligible, (a) How long does it dive last? (b) and with what speed does the person enter the water?
distance = 1/2*g*t^2
Solve for t.
velocity = g*t
Thank you so much!
To find the answers to these questions, we can use principles of physics and equations of motion. Let's break it down step by step.
(a) How long does the dive last?
To answer this, we can use the equation of motion for free fall:
\[ d = \frac{1}{2} \times g \times t^2 \]
Where:
- \( d \) is the distance fallen (in this case, the height of the cliff, which is 33m).
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s^2).
- \( t \) is the time.
Using this equation, we can rearrange it to solve for \( t \):
\[ t = \sqrt{\frac{2d}{g}} \]
Substituting the given values:
\[ t = \sqrt{\frac{2 \times 33 \, \text{m}}{9.8 \, \text{m/s}^2}} \]
Calculating:
\[ t = \sqrt{\frac{66}{9.8}} \, \text{s} \approx 2.81 \, \text{s} \]
Therefore, the dive lasts approximately 2.81 seconds.
(b) With what speed does the person enter the water?
To find the speed at which the person enters the water, we can use the equation of motion:
\[ v = g \times t \]
Where:
- \( v \) is the velocity/speed.
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s^2).
- \( t \) is the time we found in part (a), which is approximately 2.81 seconds.
Substituting the values, we have:
\[ v = 9.8 \, \text{m/s}^2 \times 2.81 \, \text{s} \]
Calculating:
\[ v = 27.5 \, \text{m/s} \]
Therefore, the person enters the water at a speed of approximately 27.5 m/s.