A ball is dropped from a height of 1.5m. how long will it take the ball to reach the ground and what velocity will it reach?

I didn't get anything of what you just wrote here.

s = u + 1/2(a) t^2

1.5 = 0m/s + 1/2 (10 or 9.81) * t^2
1.5 = 1/2 (10 or 9.81~) * t^2
√(1.5/(1/2 (10 or 9.81)))

(Given Information): d = 1.5m, Aav= 9.8 m/s^2, Vi= 0.0 m/s

d= (Vi x t) + (1/2 x Aav x t^2)

Rearrange to get

√(d/1/2 x Aav) = t
√(1.5m / 1/2 x 9.8 m/s^2) = t
0.55s = t

Well, let me put on my clown shoes and calculate this for you!

Assuming there is no air resistance (because balls really hate the wind), we can use a little physics humor to find the answers.

First, the time it takes for the ball to reach the ground can be found using the equation

t = sqrt(2h/g),

where h is the initial height (1.5m) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

So, t = sqrt(2 * 1.5 / 9.8) = sqrt(0.306) = 0.553 seconds (approximately).

Now, let's find the final velocity that the ball will reach when it hits the ground. We can use another funny equation:

v = sqrt(2gh),

where g is still the acceleration due to gravity and h is the initial height (1.5m).

So, v = sqrt(2 * 9.8 * 1.5) = sqrt(29.4) = 5.42 m/s (approximately).

So, after all my clown calculations, the ball will take approximately 0.553 seconds to reach the ground and will reach a velocity of approximately 5.42 m/s.

To determine the time it takes for the ball to reach the ground and the velocity it will reach, we can use the laws of motion and the principles of kinematics. Specifically, we can employ the equations of motion under constant acceleration.

Step 1: Identify the given information.
We are given:
- Initial height (h) = 1.5 m (the height from which the ball is dropped)
- Acceleration due to gravity (g) = 9.8 m/s^2 (assuming standard conditions)

Step 2: Determine the unknowns.
We need to find:
- Time taken for the ball to reach the ground (t)
- Velocity of the ball when it reaches the ground (v)

Step 3: Select an appropriate equation.
The first equation of motion we can use is formulated as:

h = ut + (1/2)gt^2

Where:
- h: Initial height (1.5 m)
- u: Initial velocity (which is zero as the ball is dropped)
- g: Acceleration due to gravity (9.8 m/s^2)
- t: Time taken for the ball to reach the ground

Since the initial velocity (u) is zero, the equation simplifies to:

h = (1/2)gt^2

We can rearrange this equation to isolate t:

t^2 = (2h) / g

Step 4: Calculate the time taken (t).
Substituting the given values into the equation, we have:

t^2 = (2 * 1.5) / 9.8
t^2 = 0.30612244898
t ≈ 0.554 s (taking the square root of both sides)

So, it will take approximately 0.554 seconds for the ball to reach the ground.

Step 5: Calculate the velocity (v).
We can use the second equation of motion, which is:

v = u + gt

Since the initial velocity (u) is zero, the equation simplifies to:

v = gt

Substituting the values, we have:

v = 9.8 m/s^2 * 0.554 s
v ≈ 5.4272 m/s

The ball will reach a velocity of approximately 5.4272 m/s when it reaches the ground.

x = x(initial) + v(initial)t + 1/2 at^2

0 = 1.5 + 0 + 1/2 (-10)t^2
1.5 = 5 t^2
t = 0.55 seconds

v = v(initial) + at
v = 0 + (-10)(0.55)
v = -5.5 m/s (i.e. it will reach the ground at 5.5 m/s)