A ball is thrown from a cliff at a horizontal speed of 3m/second and at the same time a second ball is dropped straight down from the cliff. How fast will the dropped ball be going after 2 seconds?

Try V = at

V = 9.8m/s*2 s = ??

WHAT IS A CHEMICAL SOLUTION AND GIVE AN EXAMPLE?

To determine the speed of the dropped ball after 2 seconds, we need to consider the effects of gravity.

The ball is dropped, so it starts from rest (zero initial velocity) and accelerates due to gravity. The acceleration due to gravity is approximately 9.8 m/s², pointing towards the ground.

Using the equation of motion: v = u + at, where
v = final velocity,
u = initial velocity,
a = acceleration, and
t = time,

we can calculate the speed of the dropped ball after 2 seconds:

u = 0 m/s (initial velocity as it is dropped)
t = 2 seconds

v = u + at
v = 0 + (9.8 m/s²)(2 seconds)
v = 0 + 19.6 m/s

Therefore, the dropped ball will be going at a speed of 19.6 m/s after 2 seconds.