Question

A student drops a ball off the top of building and records that the ball takes 3.32s to reach the ground (g = 9.8 m/s^2). What is the ball speed just before hitting the ground?​

Answer 1

Step-by-step explanation:

Here's what we know because it was given to us:

a = -9.8 m/s/s and

time = 3.32 seconds

Here's what we know because we rock physics:

v₀ = 0 (because the object was held still before it was dropped).

Here's the equation that ties all that info together in a single one-dimensional equation:

v = v₀ + at

Filling in and solving for v:

v = 0 + (-9.8)(3.32) and

v = -33m/s

The velocity is negative because the object is moving downwards and up is positive (but you knew that already too!)