If a ball is released from a window ledge,how fast will it be traveling when it reaches the ground,16 m below?

Answer 1

The formula for calculating the the distance something travels under constant acceleration for some time is (assuming no initial velocity or displacing and ignoring air resistance):

d=1/2 at^2

This formula can be used to calculate the time of the fall by putting in 16 m for "d", 9.8 m/sec/sec for acceleration due to gravity, and solving.

16m = 1/2 (9.8 m/sec/sec) t^2

16=.5*9.8 /sec/sec t^2
16/.5/9.8 sec^2 = t^2
3.26 sec^2 = t^2
1.81=t

Now that we know it took 1.8 seconds to drop we can use the formula for velocity:

v=at

v=9.8 m/sec/sec (1.81 sec)
v=17.7 m/sec

There are quicker ways to to this and formulas that are derived so you can do this calculation in one step, but I like to minimize the number of formulas to remember and work as many problems as possible with the most basic formulas.

Answer 2

Vf^2 = 2ad + Vi^2
vf^2 = 2(-9.8m/s^2)(-16m)
Vf^2 = 313.6
Vf= -17.7m/s

Answer 3

not very fast