Question

What is the total time the diver remains in the air if they spring upward with an initial speed of 8 m/s from a diving board that is 5 m above the water?

Answer 1

Approximately
`2.11\; {\rm s}`, assuming that air resistance is negligible and that
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Under the assumptions, this person would be accelerating downward at a constant
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` until reaching the water. Note that acceleration is negative since gravitational attraction points downwards.

Initial velocity of this person would be
`u = 8\; {\rm m\cdot s^(-1)}`- positive since this person was initially moving upward. The displacement during the entire motion would be
`x = (-5)\; {\rm m}`, which is negative since the water is below the initial position of the diving board.

The duration of this motion can be found in the following steps:

• Apply the SUVAT equation
  `v^(2) - u^(2) = 2\, a\, x` to find the velocity
  `v` of the person right before hitting the water.
• Divide the change in velocity by acceleration (the rate of change in velocity) to find the duration of the motion.

In the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x`:

• 
  `v` is the final velocity of the object,
• 
  `u` is the initial velocity of the object,
• 
  `a` is acceleration, and
• 
  `x` is the displacement of the object.

Rearrange this equation to find
`v` given that
`u = 8 \; {\rm m\cdot s^(-1)}`,
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`, and
`x = (-5)\; {\rm m\cdot s^(-2)}`:

`v^(2) = u^(2) + 2\, a\, x`.

Note that two different values of
`v` that would satisfy this equation: one positive, and the other negative. However, since the diver would be travelling downward right before hitting the water, the correct value of
`v` should be negative. Hence, there should be a negative sign in front of the square root:

`\begin{aligned}v &= -\sqrt{u^(2) + 2\, a\, x} \\ &= -\sqrt{8^(2) + 2\, (-9.81)\, (-5)} \; {\rm m\cdot s^(-1)} \\ &\approx (-12.732)\; {\rm m\cdot s^(-1)}\end{aligned}`.

In other words, the velocity of this person would be approximately
`(-12.732)\; {\rm m\cdot s^(-1)}` right before reaching the water.

Velocity of this person has changed from
`u = 8\; {\rm m\cdot s^(-1)}` to
`v \approx (-12.732)\; {\rm m\cdot s^(-1)}`. Divide the change in velocity by the acceleration (the rate of change in velocity) to find the duration of this motion:

`\begin{aligned}t &= (v - u)/(a) \\ &\approx ((-12.732) - (8))/((-9.81))\; {\rm s} \\ &\approx 2.11\; {\rm s}\end{aligned}`.

In other words, this person was in the air for approximately
`2.11\; {\rm s}`.