Question

practice it
use the worked example above to help you solve this problem. a diver of mass m drops from a board 16.0 m above the waters surface, as shown in the figure. neglect air resistance.
(a) use conservation of mechanical energy to find his speed 8.00 m above the waters surface.
m/s
(b) find his speed as he hits the water.
m/s
exercise
hints: getting started | im stuck!
use the values from practice it to help you work this exercise. suppose the diver vaults off the springboard, leaving it with an initial speed of 3.24 m/s upward. use energy conservation to find his speed when he strikes the water.
m/s
resources
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Explanation:

Step1: Recall mechanical - energy conservation formula

The conservation of mechanical energy is $E_{i}=E_{f}$, where $E = K+U$, $K=\frac{1}{2}mv^{2}$ and $U = mgh$. Let the water - surface be the zero - potential level ($h = 0$).

Step2: Solve part (a)

Initial state: $h_{i}=16.0\ m$, $v_{i}=0$. Final state: $h_{f}=8.00\ m$.
$E_{i}=mgh_{i}+0$ and $E_{f}=mgh_{f}+\frac{1}{2}mv_{f}^{2}$.
By $E_{i}=E_{f}$, we have $mgh_{i}=mgh_{f}+\frac{1}{2}mv_{f}^{2}$.
Cancel out the mass $m$ on both sides: $gh_{i}=gh_{f}+\frac{1}{2}v_{f}^{2}$.
$v_{f}^{2}=2g(h_{i}-h_{f})$.
Given $g = 9.8\ m/s^{2}$, $h_{i}=16.0\ m$, $h_{f}=8.00\ m$.
$v_{f}=\sqrt{2\times9.8\times(16.0 - 8.00)}=\sqrt{2\times9.8\times8.00}=\sqrt{156.8}\approx12.5\ m/s$.

Step3: Solve part (b)

Initial state: $h_{i}=16.0\ m$, $v_{i}=0$. Final state: $h_{f}=0$.
$E_{i}=mgh_{i}+0$ and $E_{f}=0+\frac{1}{2}mv_{f}^{2}$.
By $E_{i}=E_{f}$, $mgh_{i}=\frac{1}{2}mv_{f}^{2}$.
Cancel out $m$: $v_{f}=\sqrt{2gh_{i}}$.
Substitute $g = 9.8\ m/s^{2}$ and $h_{i}=16.0\ m$, $v_{f}=\sqrt{2\times9.8\times16.0}=\sqrt{313.6}\approx17.7\ m/s$.

Step4: Solve the exercise

Initial state: $h_{i}=16.0\ m$, $v_{i}=3.24\ m/s$. Final state: $h_{f}=0$.
$E_{i}=mgh_{i}+\frac{1}{2}mv_{i}^{2}$ and $E_{f}=0+\frac{1}{2}mv_{f}^{2}$.
By $E_{i}=E_{f}$, $mgh_{i}+\frac{1}{2}mv_{i}^{2}=\frac{1}{2}mv_{f}^{2}$.
Cancel out $m$: $gh_{i}+\frac{1}{2}v_{i}^{2}=\frac{1}{2}v_{f}^{2}$.
$v_{f}^{2}=2gh_{i}+v_{i}^{2}$.
Substitute $g = 9.8\ m/s^{2}$, $h_{i}=16.0\ m$, $v_{i}=3.24\ m/s$.
$v_{f}^{2}=2\times9.8\times16.0+(3.24)^{2}=313.6 + 10.4976=324.0976$.
$v_{f}=\sqrt{324.0976}\approx18.0\ m/s$.

Answer:

(a) $12.5$
(b) $17.7$
Exercise: $18.0$