Answer:
Speed = 5.87m/s
Explanation:
Given the following data;
Height of cliff = 45.1m
Horizontal distance = 17.8m
We know that acceleration due to gravity is 9.8m/²
Initial velocity = 0m/s
To find the time, we would use the second equation of motion;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Substituting into the equation, we have;
[tex] 45.1 = 0(t) + \frac {1}{2}*9.8*t^{2}[/tex]
[tex] 45.1 = 0 + 4.9*t^{2} [/tex]
[tex] 45.1 = 4.9*t^{2} [/tex]
[tex] t^{2} = \frac {45.1}{4.9} [/tex]
[tex] t = \sqrt{9.204}[/tex]
t = 3.034 secs
To find the speed;
Mathematically, speed is given by the equation;
[tex]Speed = \frac{distance}{time}[/tex]
Substituting into the above equation;
[tex]S = \frac{17.8}{3.034}[/tex]
Speed = 5.87m/s
Therefore, the cliff diver must be running at 5.87 meters per seconds before jumping off the cliff.
