A helicopter hovers H = 400 m above a lake. A bomb is…

A helicopter hovers $H = 400\ \mathrm{m}$ above a lake. A bomb is released and falls through the air (air resistance neglected). In the water it keeps moving downward with a constant deceleration of $a = 4.5\ \mathrm{m/s^2}$. It explodes on reaching the bottom, $t = 22\ \mathrm{s}$ after release. How deep is the lake?

Explanation of Method

A bomb falling through air and then water.

Step 1: The fall through the air

\[ t_1 = \sqrt{\frac{2H}{g}} = \sqrt{\frac{800}{9.81}} \approx 9.03\ \mathrm{s}, \]

\[ v_1 = \sqrt{2gH} \approx 88.6\ \mathrm{m/s}. \]

Step 2: Time in the water

\[ t_w = t - t_1 = 22 - 9.03 = 12.97\ \mathrm{s}. \]

Step 3: Depth

Starting at 88.6 m/s and slowing down at 4.5 m/s²: \[d = v_1 t_w - \tfrac12 a t_w^2 = 88.59 \cdot 12.97 - 2.25 \cdot 12.97^2 = 1149.0 - 378.5.\]

\[d \approx 770\ \mathrm{m}\]

Check

Speed at the bottom: $ 88.6 - 4.5 \cdot 12.97 \approx 30.2\ \mathrm{m/s} \gt 0 $: the bomb is still moving down when it reaches the bottom $\checkmark$

Average speed in water $ \tfrac{88.6 + 30.2}{2} = 59.4\ \mathrm{m/s} $; $ 59.4 \cdot 12.97 \approx 770\ \mathrm{m}\ \checkmark $

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $9.03$ $22$ $30.2$ $88.6$ $\text{air: } 400\ \mathrm{m}$ $\text{water: } \approx 770\ \mathrm{m}$ $+g$ $-4.5\ \mathrm{m/s^2}$ $\text{explosion}$

falling through air
sinking through water

Speed against time. In air the bomb speeds up at $g$ for 9.03 s; in the water it slows down at 4.5 m/s² for the remaining 12.97 s. The areas are the distances: 400 m of air and the depth of the lake.

Result

\[\boxed{\,d \approx 770\ \mathrm{m}\,}\]

Remark

About the wording. The statement says "acceleration of 4.5 m/s²"; like the original solution, we take it as a deceleration, since water slows the bomb down. If the bomb instead sped up at 4.5 m/s², the depth would be $1149 + 378.5 \approx 1530\ \mathrm{m}$.