Answer:
The time she takes to reach the water from when she jumps off the platform is 1.71 s
Explanation:
According to the equations of motion, we have;
v = u - g·t
v² = u² - 2·g·s
s₁ = u₁·t₁ + 1/2·g₁·t₁²
The given parameters are;
The height of the platform (assumption: above the water) = 10 m
The velocity with which she jumps, u = 3 m/s
The acceleration due to gravity, g = 9.81 m/s²
The height to which she jumps, s, is found as follows;
v² = u² - 2·g·s
At maximum height, v = 0, which gives;
0 = 3² - 2×9.81×s
2×9.81×s = 3² = 9
s = 9/(2×9.81) = 0.4587 m
s = 0.4587 m
The time to maximum height, t, is found as follows;
v = u - g·t
0 = 3 - 9.81×t
9.81×t = 3
t = 3/9.81 = 0.3058 s
The total distance, s₁ from maximum height to the water surface = s + 10 = 0.4587 + 10 = 10.4587 m = 10.46 m
The time to reach the water from maximum height, t₁, is found as follows;
s₁ = u₁·t₁ + 1/2·g₁·t₁²
Where;
s₁ = The total distance from maximum height to the water surface = 10.46 m
u₁ = The initial velocity, this time from the maximum height = 0 m/s
g₁ = The acceleration due to gravity, g (positive this time as the diver is accelerating down) = 9.81 m/s²
t₁ = The time to reach the water from maximum height
Substituting the values gives;
s₁ = u₁·t₁ + 1/2·g₁·t₁²
10.46 = 0·t₁ + 1/2·9.81·t₁²
t₁²= 10.46/(1/2×9.81) = 2.13 s²
t₁ = √2.13 = 1.46 s
Total time = t₁ + t = 1.46 + 0.3058 = 1.7066 ≈ 1.71 s.
Therefore, the time she takes to reach the water from when she jumps off the platform = 1.71 s.
