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Q27 PE

Expert-verifiedFound in: Page 261

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water with an initial speed of 15.0 m/s strikes the water with a speed of 24.8 m/s independent of the direction thrown.**

The rock will strike the water with a speed of $24.8\text{m}/\text{s}$ and is independent of the direction thrown.

**Conservation of energy: ****An isolated system's total energy is always conserved. In other words, the energy neither be created nor be destroyed; it can be only transformed from one form to another.**

Mathematically,

$\Delta \text{KE}=\Delta \text{PE}\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{f}^{2}-\frac{1}{2}m{v}_{i}^{2}=mg{h}_{f}-mg{h}_{i}$ (1.1)

Here, m is the mass of the rock, ${v}_{f}$ is the final velocity of the rock (${v}_{f}=0$ as the rock stops after striking the water),${v}_{i}$ is the initial velocity of the rock, g is the acceleration due to gravity $\left(g=9.8\text{m}/{\text{s}}^{2}\right)$, ${h}_{f}$ is the final height (${h}_{f}=0$ as the rock strikes the water), and ${h}_{i}$ is the initial height or the height of the bridge $\left({h}_{i}=20.0\text{m}\right)$.

Rearranging equation (1.1) to get an expression for the final velocity,

${v}_{f}=\sqrt{{v}_{i}^{2}+2g\left({h}_{i}-{h}_{f}\right)}$ (1.1)

When the rock is thrown downward with initial velocity will be positive i.e., ${v}_{i}=15.0\text{m}/\text{s}$. The final velocity can be calculated using equation (1.2).

Putting all known values in equation (1.2),

${v}_{f}=\sqrt{{\left(15.0\text{m}/\text{s}\right)}^{2}+2\times \left(9.8\text{m}/{\text{s}}^{2}\right)\times \left[\left(20.0\text{m}\right)-\left(0\text{m}\right)\right]}\phantom{\rule{0ex}{0ex}}=24.8\text{m}/\text{s}$

Therefore, the rock will strike the water with a speed of $24.8\text{m}/\text{s}$ when thrown downward.

When the rock is thrown downward with initial velocity will be negative i.e. ${v}_{i}=-15.0\text{m}/\text{s}$. The final velocity can be calculated using equation (1.2).

Putting all known values in equation (1.2),

${v}_{f}=\sqrt{{\left(-15.0\text{m}/\text{s}\right)}^{2}+2\times \left(9.8\text{m}/{\text{s}}^{2}\right)\times \left[\left(20.0\text{m}\right)-\left(0\text{m}\right)\right]}\phantom{\rule{0ex}{0ex}}=24.8\text{m}/\text{s}$

Therefore, the rock will strike the water with a speed of $24.8\text{m}/\text{s}$ when thrown downward.

From the above discussion it is clear that the rock will strike the water with a speed of $24.8\text{m}/\text{s}$ and is independent of the direction thrown.

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