Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m
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2m/s²
Explanation:
Given parameters:
Radius = 50m
Speed = 10m/s
Unknown:
Acceleration of the cyclist
Solution:
The acceleration of the cyclist is directed inside of the corner because his motion is inward. This is a form of centripetal acceleration;
Centripetal acceleration is given by;
a = 
v is the velocity
r is the radius
a =
= 2m/s²
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Answer:
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