A diver runs toward the edge of the cliff in Tobermory. This cliff is 44 m high (top of cliff to water level). They dive out, horizontally, at a velocity of 1.8 m/s [forward]. The sign below was posted for a reason. There is a section of rock right under the rough water extending from the cliff out for 4 m, as shown below. Does the diver hit the hidden rocks?
Answers
This question involves the concepts of the equations of motion for both the vertical motion and the horizontal motion.
The diver "does not" hit the rock.
First, we will calculate the time taken by the diver to land. For this purpose, we will use the second equation of motion for the vertical motion of the diver:
[tex]h = v_{iy}t+\frac{1}{2}gt^2[/tex]
where,
h = height of cliff = 44 m
[tex]v_{iy}[/tex] = vertical component of the initial velocity of the diver = 0 m/s
t = time taken to land = ?
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]44\ m=(0\ m/s)t+\frac{1}{2}(9.81\ m/s^2)(t)^2\\\\t = \sqrt{\frac{(44\ m)(2)}{9.81\ m/s^2}}[/tex]
t = 3 s
Now, we will calculate the horizontal distance where the diver landed. For this purpose, we will use the equation of motion for the horizontal motion of the diver. Assuming the air resistance to be zero the horizontal motion will be uniform. Hence:
[tex]s = vt\\s = (1.8\ m/s)(3\ s)\\[/tex]
s = 5.4 m
Since the section of the rock was 4 m away from the cliff. Hence, the diver crossed that rock section and safely laded at a distance of 5.4 m, which is 1.4 m ahead of the rock section.
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The attached picture shows the equations of motion for the horizontal motion and vertical motion.
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