Question

a raindrop initially falling at 34.3 m/s travels 82.0 m to the ground in free fall. what is the final velocity of the raindrop 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

We use the kinematic equation \( v_f^2 = v_i^2 + 2ad \), where \( v_i \) is the initial velocity, \( a \) is the acceleration, \( d \) is the displacement, and \( v_f \) is the final velocity. For free fall, \( a = -g=- 9.8\ m/s^2\) (negative because it's downward), \( v_i=- 34.3\ m/s\) (downward, so negative), and \( d = - 82.0\ m\) (displacement is downward, so negative).

Step2: Substitute the values into the equation

Substitute \( v_i=-34.3\ m/s \), \( a = - 9.8\ m/s^2 \), and \( d=-82.0\ m \) into \( v_f^2=v_i^2 + 2ad \):

$$
LATEXBLOCK0
$$

Step3: Solve for \( v_f \)

Take the square root of both sides. Since the velocity is downward, it will be negative:

\( v_f=-\sqrt{2783.69}\approx - 52.76\ m/s\) (we take the negative root because the direction is downward)

Answer:

\( - 52.8\) (rounded to three significant figures)