WILLL GIVE 5 STARS BRAINIEST AND THANKS AND 20 POINTS EACH ANSWER A diver performs a dive from a 10 m-platform. She
jumps vertically upwards at 3 m s initially.
Find the time she takes to reach the water surface

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.

f=ma newton's second law 7x8=56n