Question

breaking down 2-d motion
vectors and 2-d kinematics
a ball initially at rest is dropped
from the roof of a 42.5 m tall
building and travels in free fall.
what is the final velocity of the ball
just before it hits the ground?
$v_f = ? \\ m/s$
do not account for air resistance.
remember, downward velocity is a negative vector (-)

Explanation:

Step1: Identify the kinematic equation

For free - fall motion (constant acceleration \(a = g=- 9.8\ m/s^{2}\) (taking downwards as negative), initial velocity \(v_{i}=0\ m/s\) (since the ball is dropped from rest), displacement \(d=- 42.5\ m\) (negative because it is in the downward direction), we use the kinematic equation \(v_{f}^{2}=v_{i}^{2}+2ad\)

Step2: Substitute the values into the equation

We know that \(v_{i} = 0\ m/s\), \(a=-9.8\ m/s^{2}\), and \(d = - 42.5\ m\). Substituting these values into the equation \(v_{f}^{2}=v_{i}^{2}+2ad\), we get:
\(v_{f}^{2}=0^{2}+2\times(-9.8)\times(- 42.5)\)
First, calculate \(2\times(-9.8)\times(-42.5)=2\times9.8\times42.5 = 19.6\times42.5=833\)
So \(v_{f}^{2}=833\)

Step3: Solve for \(v_{f}\)

Take the square root of both sides. Since the velocity is downward (negative direction), we take the negative square root:
\(v_{f}=-\sqrt{833}\approx - 28.9\ m/s\) (We can also calculate \(\sqrt{833}\approx28.9\), and since the direction is downward, the velocity is \(- 28.9\ m/s\))

Answer:

\(-28.9\) (or approximately \(-29\) depending on the level of precision, but using \(g = 9.8\ m/s^{2}\), the more accurate value is approximately \(-28.9\))