Answer:
A) 0.9844 s
B) x2 = 0.4587 m
C) v = 6.657 m/s
Explanation:
We are given;
Height of take off point above pool; x1 = 1.8 m
Initial take off velocity; u = 3 m/s
Final velocity at highest point before free fall; v = 0 m/s
B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.
Thus, we will be using equation of motion and we have;
v² = u² + 2gs
Now, let s = x2 which will be the distance between take off and the top before free fall.
So;
v² = u² + 2g(x2)
Now,since the motion is against gravity, g will be negative.
Thus;
v² = u² + 2(-9.81)(x2)
Plugging in the relevant values to give;
0² = 3² - (19.62x2)
19.62(x2) = 9
x2 = 9/19.62
x2 = 0.4587 m
A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.
Thus, using equation of motion;
v = u + gt
Again, g = -9.81
Thus;
0 = 3 - 9.81t1
9.81t1 = 3
t1 = 3/9.81
t1 = 0.3058 s
Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;
s = ut + ½gt²
Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m
And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²
Thus;
2.2587 = 0 - ½(-9.81)(t2)²
2.2587 = 4.905(t2)²
(t2)² = 2.2587/4.905
(t2)² = 0.4605
t2 = √0.4605
t2 = 0.6786 s
Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s
C) Velocity when feet hit the water would be given by;
v = u + gt
Where u = 0 m/s and t = t2 = 0.6786
Since it's in direction of gravity, g = 9.81 m/s
v = 0 + (0.6786 × 9.81)
v = 6.657 m/s
:
, and the two locations have potential of 
and 
, therefore the work is