Question

breaking down 2-d motion
vectors and 2-d kinematics
an egg initially at rest is dropped
from the roof of a 31.5 m tall
building and travels in free fall.
what is the final velocity of the egg
just before it hits the ground?
$v_f = ? \\, \text{m/s}$
do not account for air resistance.
remember, downward velocity is a negative vector (-).

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation for free fall: \( v_f^2 = v_i^2 + 2gh \), where \( v_i = 0 \) (initial velocity, at rest), \( g = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity, negative because downward), and \( h = -31.5 \, \text{m} \) (displacement, negative because downward).

Step2: Substitute values into the equation

Substitute \( v_i = 0 \), \( g = -9.8 \, \text{m/s}^2 \), and \( h = -31.5 \, \text{m} \) into \( v_f^2 = v_i^2 + 2gh \):
\( v_f^2 = 0^2 + 2(-9.8)(-31.5) \)
Calculate the right-hand side: \( 2 \times 9.8 \times 31.5 = 617.4 \)

Step3: Solve for \( v_f \)

Take the square root of both sides. Since the velocity is downward, it will be negative:
\( v_f = -\sqrt{617.4} \approx -24.85 \, \text{m/s} \) (we can also use \( g = 9.8 \) and consider direction; alternatively, if we take \( g = 9.8 \) and \( h = 31.5 \) with direction in mind, \( v_f = -\sqrt{2gh} = -\sqrt{2 \times 9.8 \times 31.5} = -\sqrt{617.4} \approx -24.85 \), or more accurately, using \( g = 9.81 \) for better precision: \( \sqrt{2 \times 9.81 \times 31.5} \approx \sqrt{618.03} \approx 24.86 \), so \( v_f \approx -24.9 \) or -24.8 m/s. But let's use \( g = 9.8 \))

Wait, let's recalculate with \( g = 9.8 \):

\( 2 \times 9.8 \times 31.5 = 2 \times 9.8 = 19.6; 19.6 \times 31.5 = 617.4 \)

\( \sqrt{617.4} \approx 24.85 \), so \( v_f \approx -24.85 \, \text{m/s} \), which we can round to -24.9 or -24.8. But maybe the problem expects using \( g = 9.8 \) and getting approximately -24.8 or -24.9. Alternatively, let's check the calculation again.

Wait, the displacement \( h \) is -31.5 m (if upward is positive), initial velocity \( v_i = 0 \), acceleration \( a = -9.8 \, \text{m/s}^2 \). The kinematic equation \( v_f^2 = v_i^2 + 2ah \)

So \( v_f^2 = 0 + 2(-9.8)(-31.5) = 2 \times 9.8 \times 31.5 = 617.4 \)

Then \( v_f = -\sqrt{617.4} \approx -24.85 \, \text{m/s} \), so approximately -24.9 m/s or -24.8 m/s. Let's compute \( \sqrt{617.4} \):

\( 24^2 = 576 \), \( 25^2 = 625 \), so 24.8^2 = (24 + 0.8)^2 = 24^2 + 2240.8 + 0.8^2 = 576 + 38.4 + 0.64 = 615.04

24.85^2 = (24.8 + 0.05)^2 = 24.8^2 + 224.80.05 + 0.05^2 = 615.04 + 2.48 + 0.0025 = 617.5225, which is very close to 617.4. So \( \sqrt{617.4} \approx 24.85 - \) a little, so approximately 24.84, so \( v_f \approx -24.8 \, \text{m/s} \) or -24.9 m/s. Depending on the precision, but let's use \( g = 9.8 \) and calculate:

\( 2 * 9.8 * 31.5 = 617.4 \)

\( \sqrt{617.4} \approx 24.85 \), so \( v_f \approx -24.9 \, \text{m/s} \) (rounded to one decimal place) or -24.8 m/s.

Alternatively, maybe the problem expects using \( g = 10 \) for approximation, but no, standard is 9.8.

Wait, let's do it with \( g = 9.8 \):

\( v_f = -\sqrt{2gh} = -\sqrt{2 * 9.8 * 31.5} = -\sqrt{617.4} \approx -24.85 \, \text{m/s} \), so the final velocity is approximately -24.9 m/s (or -24.8 m/s, depending on rounding).

Answer:

\( \boxed{-24.9} \) (or -24.8, depending on calculation precision; using \( g = 9.8 \), the exact value is approximately -24.85, which can be rounded to -24.9 or -24.8)