You throw a ball downward from a window at a speed of 2.0m/s. The ball accelerates at 9.8m/s^2. How fast is it moving when it hits the sidewalk 2.5m below?

initial speed: 2 m/s

a: -9.8 m/s^2
s: -2.5 m
final speed: ?
Vf^2 - Vi^2 = 2as

v^2 - 2^2 = 2•-9.8•-2.5

v^2-4 = 49

v^2 = 53
final speed is approx. 7.28

Well, well, well, looks like the ball is ready for a high-speed trip to the sidewalk! Let's calculate its velocity when it reaches the ground.

We know the initial velocity of the ball is 2.0 m/s and the acceleration is 9.8 m/s^2. Since the ball is going down, we take the acceleration value as a positive value.

Now, to find the final velocity, we can use the good old kinematic equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

Plugging in the values:

v^2 = 2.0^2 + 2(9.8)(-2.5)

v^2 = 4 + (-49)(-2.5)

Now, let's do some math magic:

v^2 = 4 + 122.5

v^2 = 126.5

Taking the square root of both sides, we have:

v = √126.5

v ≈ 11.25 m/s

So clowning around aside, when the ball hits the sidewalk 2.5 meters below, it'll be moving at approximately 11.25 m/s. Keep an eye out!

To find the final speed of the ball when it hits the sidewalk, we can use the kinematic equation:

vf^2 = vi^2 + 2aΔx

Where:
vf = final velocity
vi = initial velocity
a = acceleration
Δx = change in position (displacement)

Given:
vi = 2.0 m/s (initial velocity)
a = 9.8 m/s^2 (acceleration)
Δx = 2.5 m (change in position)

Let's substitute these values into the kinematic equation:

vf^2 = (2.0 m/s)^2 + 2(9.8 m/s^2)(2.5 m)

Simplifying:

vf^2 = 4.0 m^2/s^2 + 2(9.8 m/s^2)(2.5 m)

vf^2 = 4.0 m^2/s^2 + 49 m^2/s^2

vf^2 = 53 m^2/s^2

Taking the square root of both sides to solve for vf:

vf ≈ √53 m/s

Therefore, the ball is moving at approximately 7.28 m/s when it hits the sidewalk.

To calculate the final velocity of the ball when it hits the sidewalk, we can use the equation of motion:

v^2 - u^2 = 2 * a * s

where:
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance

Given:
u = 2.0 m/s (initial velocity)
a = 9.8 m/s^2 (acceleration)
s = 2.5 m (distance)

Let's solve for v.

v^2 - u^2 = 2 * a * s

v^2 = u^2 + 2 * a * s

v^2 = (2.0 m/s)^2 + 2 * (9.8 m/s^2) * (2.5 m)

v^2 = 4.0 m^2/s^2 + 49.0 m^2/s^2

v^2 = 53.0 m^2/s^2

v = √53.0 m/s

v ≈ 7.280 m/s (rounded to three decimal places)

Therefore, when the ball hits the sidewalk 2.5m below, it will be moving at approximately 7.280 m/s.