Answer:
(a) 0
(b) 10ML
(c)
![10ML(1 - cos(\theta))](https://img.qammunity.org/2020/formulas/physics/college/hr5uzwvkh0m7drrrm00gbmp1cpva7rw8qn.png)
(d)
![10ML(1 + sin(\phi))](https://img.qammunity.org/2020/formulas/physics/college/tf64fyrr38w8jgsdyfk5tstmehp5jf06wu.png)
Step-by-step explanation:
(a) When hanging straight down. The child is at the lowest position. His potential energy with respect to this point would also be 0.
(b) Since the rope has length L m. When the rope is horizontal, he is at L (m) high with respect to the lowest swinging position. His potential energy with respect to this point should be
![E_h = mgh = 10ML](https://img.qammunity.org/2020/formulas/physics/college/h4170tpshu1aj43j90axxdfn6471axvlzv.png)
where g = 10m/s2 is the gravitational acceleration.
(c) At angle
from the vertical. Vertically speaking, the child should be at a distance of
to the swinging point, and a vertical distance of
to the lowest position. His potential energy to this point would be:
![E_(\theta) = mgh = 10M(L - Lcos(\theta)) = 10ML(1 - cos(\theta))](https://img.qammunity.org/2020/formulas/physics/college/f3pwfahl4ilf3b280bq91wd5j9wplcgmda.png)
(d) at angle
from the horizontal. Suppose he is higher than the horizontal line. This would mean he's at a vertical distance of
from the swinging point and higher than it. Therefore his vertical distance to the lowest point is
![L + Lsin(\phi) = L(1 + sin(\phi))](https://img.qammunity.org/2020/formulas/physics/college/j03pp56quifvz3sen2yct9cdq59dcfefwn.png)
His potential energy to his point would be:
![E_(\phi) = mgh = 10ML(1 + sin(\phi))](https://img.qammunity.org/2020/formulas/physics/college/1il62g4gkez9wes7th76pcc4rpjbe1mm2o.png)