A baseball is seen to pass upward by a window 27 m above the street with a vertical speed of 12 m/s. The ball was thrown from the street.

(a) What was its initial speed?
(b) What altitude does it reach?
(c) How long after it was thrown did it pass the window?
(d) After how many more seconds does it reach the street again?

To solve this problem, we can use the equations of motion. Let's break down each part of the problem step-by-step.

(a) What was its initial speed?
The initial speed of the baseball can be found using the equation for vertical velocity (v) and time (t):
v = u + gt

Given:
v = 12 m/s (vertical speed upwards)
g = 9.8 m/s² (acceleration due to gravity)

Rearranging the equation, we have:
u = v - gt

Substituting the values, we get:
u = 12 m/s - (9.8 m/s²)(0) = 12 m/s

Therefore, the initial speed of the baseball is 12 m/s.

(b) What altitude does it reach?
To find the altitude the ball reaches, we can use the equation for displacement (s) in vertical motion:
s = ut + (1/2)gt²

Given:
u = 12 m/s (initial velocity)
g = 9.8 m/s² (acceleration due to gravity)
t = ? (time taken to reach the altitude)

The initial position is at the street level, so s = 27 m.

Rearrange the equation:
t² - (2u/g)t + (2s/g) = 0

Substituting the given values, we have:
t² - (2*12/9.8)t + (2*27/9.8) = 0

Solving the quadratic equation using the quadratic formula or factoring, we find:
t ≈ 2.75 seconds or t ≈ -2.75 seconds

Since time cannot be negative in this context, the ball reaches the altitude in approximately 2.75 seconds.

(c) How long after it was thrown did it pass the window?
The time for the ball to reach the window can be calculated using the equation:
s = ut + (1/2)gt²

Given:
u = 12 m/s (initial velocity)
g = 9.8 m/s² (acceleration due to gravity)
t = ? (time taken to pass the window)
s = 27 m (altitude of the window)

Rearrange the equation:
t² - (2u/g)t + (2s/g) = 0

Substitute the given values:
t² - (2*12/9.8)t + (2*27/9.8) = 0

Solving the quadratic equation, we find:
t ≈ 3.29 seconds or t ≈ -0.18 seconds

Since time cannot be negative in this context, the ball passes the window approximately 3.29 seconds after it was thrown.

(d) After how many more seconds does it reach the street again?
Since the ball reaches its maximum altitude in approximately 2.75 seconds, it will take the same amount of time to come back down and reach the street.

Therefore, after approximately 2.75 seconds, the ball will reach the street again.

To solve this problem, we can use the equations of motion for an object in free fall and the principles of kinematics. Let's break it down step by step:

(a) What was its initial speed?
To find the initial speed of the baseball, we need to consider the vertical motion only since there is no mention of any horizontal motion. The initial vertical velocity, v₀, can be found using the formula:

v = v₀ + at

Where v is the final vertical velocity (12 m/s), a is the acceleration due to gravity (which is approximately -9.8 m/s²), and t is the time taken. Since the baseball starts from rest, the initial vertical velocity can be calculated as:

v₀ = v - at

Substituting the values, we get:

v₀ = 12 m/s - (-9.8 m/s²) * t, where t = 0 (as the ball starts from rest)

v₀ = 12 m/s

Therefore, the initial speed of the baseball is 12 m/s.

(b) What altitude does it reach?
To find the altitude reached by the baseball, we can use the equation for vertical displacement, assuming the ball reaches its maximum height. The equation is:

Δy = v₀t + (1/2)at²

Where Δy is the vertical displacement, v₀ is the initial vertical velocity, a is the acceleration due to gravity, and t is the time taken.

At the maximum height, the vertical velocity becomes zero, so we can use the equation:

0 = v₀ - g * t, where g is the acceleration due to gravity

Solving for t:

t = v₀ / g

Substituting the given values:

t = 12 m/s / 9.8 m/s² ≈ 1.22 s

Now, substituting the values of v₀ and t into the equation for displacement:

Δy = (12 m/s)(1.22 s) + (1/2)(-9.8 m/s²)(1.22 s)²

Δy ≈ 7.32 m

Therefore, the baseball reaches an altitude of approximately 7.32 m.

(c) How long after it was thrown did it pass the window?
To find the time it takes for the baseball to pass the window, we need to use the equation for vertical displacement again. Since the initial vertical velocity was 0, the equation becomes:

Δy = (1/2)at²

Substituting the values:

27 m = (1/2)(-9.8 m/s²)t²

Solving for t:

t² = (2 * 27 m) / -9.8 m/s²

t ≈ √(-5.51 s²)

As time cannot be negative in this context, the square root of a negative value implies that the ball did not reach the window on its way up. Therefore, we need to calculate the time it takes for the ball to fall from the maximum height to the window.

Using the equation for displacement:

Δy = (1/2)at²

Substituting the values:

27 m = (1/2)(9.8 m/s²)t²

Solving for t:

t² = (2 * 27 m) / 9.8 m/s²

t ≈ √(5.51 s²)

Therefore, it takes approximately 2.35 seconds for the ball to pass the window after it was thrown.

(d) After how many more seconds does it reach the street again?
To find the time it takes for the ball to reach the street again, we can use the equation for vertical displacement:

Δy = v₀t + (1/2)at²

Since the ball returns to the same vertical displacement as it started, the equation becomes:

0 = v₀t + (1/2)at²

We need to solve for t in this equation. Let's substitute the known values:

0 = 12 m/s * t + (1/2)(-9.8 m/s²)t²

Rearranging the equation:

(1/2)(-9.8 m/s²)t² + 12 m/s * t = 0

Using factoring, we can rewrite this equation:

t[(1/2)(-9.8 m/s²) * t + 12 m/s] = 0

From this equation, we have two possible solutions for t:

t₁ = 0 (initial time when the ball is thrown)
t₂ = (√(12 m/s) - √[-9.8 m/s²)^2 - 4(1/2)(-9.8 m/s²)(0)]) / 2 * (1/2)(-9.8 m/s²)

t₂ ≈ (√(12 m/s) - √(0)) / (1/2)(-9.8 m/s²)

t₂ ≈ (√(12 m/s)) / (1/2)(-9.8 m/s²)

t₂ ≈ (√(12 m/s)) / (-4.9 m/s²)

t₂ ≈ (3.464 s) / (-4.9 m/s²)

t₂ ≈ -0.707 s

Since time cannot be negative in this context, the negative value is not valid. Therefore, the ball reaches the street again after approximately 0.707 seconds.

why do communities need laws?

x = 1/2 a t^2 + v0 t + x0

and v = at + v0

(a)needs no translation, but some work: with v=12m/s, solve the velocity equation for the special time t (call it t) when the ball is at the window (in terms of 12m/sec and v0). Don't lose your nerve. Now plug that expression for the t1 into the x equation, with x equal to 27 m, and solve for v0.
(b) find the time, t2, when v=0 (That's at the top of the path.) plug that time into the position equation to find the x value at the highest point.
(c)Now that you know v0, return to the velocity equation in part a and solve for t.
(d)Use the position equation and ask "what time is it when x = 0?"