A pool ball leaves a 0.6 m high table with an initial horizontal velocity of 2.4 m.s.calculate horizontal and vertical velocity and time taken for the pool ball to fall to the ground

two problems connected only by falling time

vertical problem:
Vi = initial y velocity = 0
v = Vi - g t
h = Hi + Vi t - .5 g t^2
so at ground
0 = 0.6 + 0 - 4.9 t^2
t^2 = 0.6/4.9
t = 0.35 s to fall

v at ground = 9.81* 0.35

horizontal velocity = 2.4 until it hits

range = 2.4 * .35

To calculate the horizontal and vertical velocity of the pool ball, as well as the time taken for it to fall to the ground, we can use the equations of motion.

Given:
Initial vertical displacement, h = 0.6 m
Initial horizontal velocity, u = 2.4 m/s
Acceleration due to gravity, g = 9.8 m/s^2

1. Calculate the vertical velocity (v) using the equation:
v^2 = u^2 + 2gh

Rearranging the equation, we have:
v = sqrt(u^2 + 2gh)

Substituting the known values into the equation:
v = sqrt((2.4 m/s)^2 + 2 * 9.8 m/s^2 * 0.6 m)
v = sqrt(5.76 m^2/s^2 + 11.76 m^2/s^2)
v = sqrt(17.52 m^2/s^2)
v = 4.18 m/s

So, the vertical velocity of the pool ball is 4.18 m/s.

2. Calculate the horizontal velocity (v_h) by considering that there is no horizontal force acting on the ball, so the horizontal velocity remains constant throughout the motion. Hence, v_h = u.

Therefore, the horizontal velocity of the pool ball is 2.4 m/s.

3. Calculate the time taken (t) for the pool ball to fall to the ground using the equation:
t = (v - u)/g

Substituting the known values:
t = (4.18 m/s - 0 m/s) / 9.8 m/s^2
t = 4.18 s / 9.8 m/s^2
t = 0.4265 s (rounded to four decimal places)

So, the time taken for the pool ball to fall to the ground is approximately 0.4265 seconds.

To calculate the horizontal and vertical velocity, as well as the time taken for the pool ball to fall to the ground, we can use the equations of motion.

First, let's consider the vertical motion of the pool ball. We can use the equation for vertical displacement (Δy) during free fall:

Δy = (V₀y * t) + (0.5 * a * t²),

where:
Δy is the vertical displacement (0.6 m in this case),
V₀y is the initial vertical velocity (we need to calculate this),
t is the time taken (we need to calculate this), and
a is the acceleration due to gravity (-9.8 m/s²).

Since the ball is leaving the table horizontally, its initial vertical velocity, V₀y, is 0 m/s.

Substituting the values into the equation, we get:

0.6 = (0 * t) + (0.5 * (-9.8) * t²).

Simplifying the equation gives us:

0.6 = -4.9t².

Now, we can solve for t. Rearranging the equation gives us:

t² = 0.6 / -4.9.

t² ≈ -0.12245.

Since time cannot be negative, we discard the negative sign and calculate the square root of the positive value:

t ≈ √0.12245.

t ≈ 0.35 seconds (rounded to two decimal places).

Now let's calculate the horizontal velocity. The horizontal velocity remains constant throughout the motion since there are no horizontal forces acting on the pool ball. Therefore, the initial horizontal velocity, V₀x, is equal to the given value of 2.4 m/s.

So, the horizontal velocity is 2.4 m/s.

To summarize:
- The horizontal velocity is 2.4 m/s.
- The initial vertical velocity is 0 m/s.
- The time taken for the pool ball to fall to the ground is approximately 0.35 seconds.