A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2 s later. How high
was the cliff and how far from its base did the diver hit the water?
work needed
work needed
1 answer:
Refer to the diagram shown below.
h = the height of the cliff.
d = the distance of the landing spot from the base of the cliff.
Assume that air resistance may be neglected, and g = 9.8 m/s².
The horizontal distance traveled in 2 seconds is
d = (1.8 m/s)*(2 s) = 3.6 m
The initial vertical velocity component is zero.
Therefore the vertical height traveled before landing on the water obeys the equation
h = (1/2)*(9.8 m/s²)*(2 s)² = 19.6 m
Answers:
The height of the cliff is 19.6 m
The distance of the landing spot from the base of the cliff is 3.6 m
h = the height of the cliff.
d = the distance of the landing spot from the base of the cliff.
Assume that air resistance may be neglected, and g = 9.8 m/s².
The horizontal distance traveled in 2 seconds is
d = (1.8 m/s)*(2 s) = 3.6 m
The initial vertical velocity component is zero.
Therefore the vertical height traveled before landing on the water obeys the equation
h = (1/2)*(9.8 m/s²)*(2 s)² = 19.6 m
Answers:
The height of the cliff is 19.6 m
The distance of the landing spot from the base of the cliff is 3.6 m
You might be interested in
Answer:
Step-by-step explanation:
Let third cook take x hours to prepare the same number of pies y alone
Let y be the number of pies
In 1 hour ,third cook prepare pies=
One experienced cook takes time to prepare enough pies =4 hours
In 1 hour , cook prepare pies=
Another cook takes time to prepare same number of pies =7 hours
In 1 hour , another cook prepare pies=
All three cooks take time to prepare the same number of pies=2 hours
In 1 hour,all three cooks prepare pies=
According to question
hours
Hence, the third cook takes time to prepare the same number of pies alone=