# If an object is dropped, how fast will it be moving after 16 s?

Apr 11, 2018

Theoretical:
$v = u + a t$, where:

• $v$ = final velocity ($m {s}^{-} 1$)
• $u$ =initial velocity ($m {s}^{-} 1$)
• $a$ = acceleration ($m {s}^{-} 2$)
• $t$ = time ($s$)

We will take $a = 9.81 m {s}^{-} 2$

$v = 0 + 16 \left(9.81\right) = 156.96 m {s}^{-} 1 \approx 157 m {s}^{-} 1$

Realistic:
The speed will depend on the shape of the object and surface area (large drag force or small drag force), height it is dropped from (to allow for a 16s fall), environment (different mediums will have different drag forces for the same object), how high the object is (higher up you go, the smaller the drag force but the smaller the acceleration due to gravity).