Question

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-like nature of the board to help propel him into the air. Assume that the diver’s motion is essentially vertical. He leaves the board, which is 3.0 m above the water, with a speed of 6.3 m/s. How long is the diver in the air, from the moment he leaves the board until

Answer 1

Answer:

The diver is in the air for $1.65s$.

Explanation:

Hi

• Known data $v_{i}=6.3m/s, y_{i}=3.0m$ and $g=9.8m/s^{2}$.

1. We need to find the time when the diver reaches the highest point, so we use the following equation $v_{f} =v_{i}-gt$ with $v_{f}=0ms$ so $t_{1}=\frac{v_{i}-v_{f} }{g}=\frac{6.3m/s}{9.8m/s^{2} }=0.64 s$.
2. Then we need to find the highest point, so we use $y=v_{i}t-\frac{1}{2} gt^{2}=(6.3m/s)(0.64s)-\frac{1}{2} (9.8m/s^{2})(0.64s)^{2}=2.03m$, this is above the springboard so the highest point is $y_{max}=5.03m$.

Finally, we need to find the time in free fall, so we use $y_{f}=y_{i}+v_{i}t-\frac{1}{2}gt^{2}$, at this stage $v_{i}=0m/s, y_{i}=5.03m$ and $y_{f}=0m$, therefore we have $0m=5.03-\frac{1}{2}(9.8m/s^{2})t^{2}$, and solving for $t_{2}=\sqrt{\frac{5.03m}{4.9m/s^{2}}} =\sqrt{1.02s^{2}}=1.01s$.

Last steep is to sum $t_{1}$ and $t_{2}$, so $t_{T}=t_{1}+t_{2}=0.64s+1.01s=1.65s$.

Answer 2

The total time spent by the diver in the air from the moment he leaves the board until he gets to the water is 1.65 s

We'll begin by calculating the time taken to get to the maximum height from the board.

Initial velocity (u) = 6.3 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Time to reach maximum height (t₁) =?

v = u – gt (since the diver is going against gravity)

0 = 6.3 – 9.8t₁

Collect like terms

0 – 6.3 = –9.8t₁

–6.3 = –9.8t₁

Divide both side by –9.8

t₁ = –6.3 / –9.8

t₁ = 0.64 s

Next, we shall determine the maximum height reached by the diver from the board

Initial velocity (u) = 6.3 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Maximum Height from the board (h₁) =?

v² = u² – 2gh (since the diver is going against gravity)

0² = 6.3² – (2 × 9.8 × h₁)

0 = 39.69 – 19.6h₁

Collect like terms

0 – 39.69 = –19.6h₁

–39.69 = –19.6h₁

Divide both side by –19.6

h₁ = –39.69 / –19.6

h₁ = 2.03 m

Next, we shall determine the height from the maximum height reached by the diver to the water.

Maximum height from the board (h₁) = 2.03 m

Height of board from water (h₂) = 3 m

Height of diver from maximum height to water (H) =?

H = h₁ + h₂

H = 2.03 + 3

H = 5.03 m

Next, we shall determine the time taken by the diver to fall from the maximum height to the water.

Height (H) = 5.03 m

Acceleration due to gravity (g) = 9.8 m/s²

Time to fall from maximum height to water (t₂) =?

H = ½gt²

5.03 = ½ × 9.8 × t₂²

5.03 = 4.9 × t₂²

Divide both side by 4.9

t₂² = 5.03 / 4.9

Take the square root of both side

t₂ = √(5.03 / 4.9)

t₂ = 1.01 s

Finally, we shall determine the total time spent by the diver in the air.

Time to reach maximum height (t₁) = 0.64 s

Time to fall from maximum height to water (t₂) = 1.01 s

Total time in air (T) =?

T = t₁ + t₂

T = 0.64 + 1.01

T = 1.65 s

Therefore, the total time spent by the diver in the air is 1.65 s

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