A crazed physics student goes to the roof of the school 14.2 meters above the ground and drops a pumpkin straight down with an initial velocity of 0 m/s how long does it take for the pumpkin to hit the ground

To solve this problem, we use acceleration due to gravity (g), which we will assume, this close to the Earth, is 9.8 m/s/s. To find time, we look at four known kinematic equations:

d = (Vf-Vi/2) t

d=Vit + 1/2at^2

Vf^2 = Vi^2 + 2ad

Vf = Vi + at

With the given information, which is a displacement of 14.2 meters downward (-14.2), an initial velocity of 0, and acceleration due to gravity downwards (-9.8 meters per second per second), we could solve the problem by plugging in these constants for our second equation and solving for t. However, to avoid the inefficiency of possibly using the quadratic formula, we can instead solve for final velocity, and use that to find our change in time. If we look at our third equation, we can simplify it to:

Vf = Square Root Of; 2 (-9.8) (-14.2)

Once you have a final velocity, solving for time becomes much easier, where

Vf = Vi + at,

or rearranged as

t = (Vf-Vi) / a