Question

1. A diver jumps from a 13 m tower, with no initial velocity.

Answer 1

Answer:

Assuming that air resistance is negligible.

a. Approximately $16.1\; {\rm m\cdot s^{-1}}$.

b. Approximately $19.0\; {\rm m \cdot s^{-1}}$.

Explanation:

Let $m$ denote the mass of this diver.

If the initial speed of the diver is $v_{0}$, the initial kinetic energy ($\text{KE}$) of this diver would be $(1/2)\, m \, {v_{0}}^{2}$.

If the height of this diver is $h$, the gravitational potential energy ($\text{GPE}$) of this diver would be $m\, g \, h$.

The initial mechanical energy of this diver (sum of $\text{KE}$ and $\text{GPE}$) would thus be: $(((1/2)\, m\, {v_{0}}^{2}) + (m\, g\, h}))$.

If air resistance on the diver is negligible, the mechanical energy of this diver would stay the same until right before the diver impacts the water. The entirety of the initial mechanical energy, $(((1/2)\, m\, {v_{0}}^{2}) + (m\, g\, h}))$, would be converted to kinetic energy by the time of impact.

Rearrange the equation $\text{KE} = (1/2)\, m\, v^{2}$ to find an expression for the speed of the diver:

$\begin{aligned} v = \sqrt{\frac{2\, \text{KE}}{m}}\end{aligned}$.

Thus, if the kinetic energy of the diver is $(((1/2)\, m\, {v_{0}}^{2}) + (m\, g\, h}))$, the speed of the diver would be:

$\begin{aligned} v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{2\times \frac{(1/2)\, m\, {v_{0}}^{2} + (m\, g\, h)}{m}} \\ &= \sqrt{2 \times ((1/2)\, {v_{0}}^{2}) + (g\, h))} \\ &= \sqrt{{v_{0}}^{2} + 2\, g\, h}\end{aligned}$.

Notice how $m$, the mass of the diver was eliminated from the expression.

If the diver started with no initial speed ($v_{0} = 0\; {\rm m\cdot s^{-1}}$) at a height of $h = 13\; {\rm m}$, the speed of the diver right before impact with water would:

$\begin{aligned} v &= \sqrt{{v_{0}}^{2} + 2\, g\, h} \\ &= \sqrt{{(0\; {\rm m\cdot s^{-1}})}^{2} + (2 \times 10\; {\rm m\cdot s^{-2} \times 13\; {\rm m})} \\ &\approx 16.1\; {\rm m\cdot s^{-1}}\end{aligned}$.

If the diver started with an initial velocity of $10\; {\rm m\cdot s^{-1}}$ upwards (initial speed $v_{0} = 10\; {\rm m\cdot s^{-1}}$) from a height of $h = 13\; {\rm m}$, the speed of the diver right before impact with water would be:

$\begin{aligned} v &= \sqrt{{v_{0}}^{2} + 2\, g\, h} \\ &= \sqrt{{(10\; {\rm m\cdot s^{-1}})}^{2} + (2 \times 10\; {\rm m\cdot s^{-2} \times 13\; {\rm m})} \\ &\approx 19.0 \; {\rm m\cdot s^{-1}}\end{aligned}$.