Question

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.

Short answer

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

Step-by-step solution

Conservation of energy

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,

$$\begin{gathered} \Delta \text{KE}=\Delta \text{PE} \\ \frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=mg{h}_f-mg{h}_i \end{gathered}$$ (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 $\left(h_i=20.0\text{m}\right)$.

Proof that the rock will strike the water with a speed of  when thrown downward

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),

$$\begin{gathered} v_f=\sqrt{{\left(15.0\text{m}/\text{s}\right)}^2+2\times \left(9.8\text{m}/{\text{s}}^2\right)\times \left[\left(20.0\text{m}\right)-\left(0\text{m}\right)\right]} \\ =24.8\text{m}/\text{s} \end{gathered}$$

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

Proof that the rock is independent of the direction thrown

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),

$$\begin{gathered} v_f=\sqrt{{\left(-15.0\text{m}/\text{s}\right)}^2+2\times \left(9.8\text{m}/{\text{s}}^2\right)\times \left[\left(20.0\text{m}\right)-\left(0\text{m}\right)\right]} \\ =24.8\text{m}/\text{s} \end{gathered}$$

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.