A. A baseball is seen to pass upward by a window 28.1 m above the street with a vertical speed of 13.7 m/s. If the ball was thrown from the street, what was its initial speed?
B. What altitude does it reach?
C. How long before the ball passed the window was it thrown?
D. How long after the baseball passed the window does it reach the street again?
A. Total energy per mass = Vo^2/2
= (13.7)^2/2 + g*28.1
Solve for initial speed, Vo
B. Vo^2/2 = g*Hmax
Solve for max altitude, Hmax
c. (Velocity change)/g = time going up
= (Vo - 13.7)/g
d. (time spent going up after passing window) + (fall time)
= 13.7/g + Vo/g
A. To find the initial speed of the baseball, we can use the equation for vertical motion:
v^2 = u^2 + 2as
Where:
v = final velocity (13.7 m/s, as it is moving upward)
u = initial velocity (what we need to find)
a = acceleration due to gravity (-9.8 m/s^2, as it acts downward)
s = displacement (28.1 m, as the ball is 28.1 m above the street)
By substituting the values into the equation, we can solve for u:
(13.7)^2 = u^2 + 2(-9.8)(28.1)
187.69 = u^2 - 549.76
u^2 = 737.45
Taking the square root of both sides:
u ≈ 27.12 m/s
Therefore, the initial speed of the baseball was approximately 27.12 m/s.
B. To find the altitude the baseball reaches, we can use the equation for vertical motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s when the ball reaches its highest point)
u = initial velocity (27.12 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
s = altitude (what we need to find)
By substituting the values into the equation, we can solve for s:
0 = (27.12)^2 + 2(-9.8)s
0 = 734.98 - 19.6s
19.6s = 734.98
s ≈ 37.51 m
Therefore, the baseball reaches an altitude of approximately 37.51 m.
C. To find how long before the ball passed the window it was thrown, we can use the equation for displacement in vertical motion:
s = ut + (1/2)at^2
Where:
s = displacement (28.1 m)
u = initial velocity (27.12 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (what we need to find)
By substituting the values into the equation, we can solve for t:
28.1 = 27.12t + (1/2)(-9.8)t^2
0 = -4.9t^2 + 27.12t - 28.1
Using the quadratic formula:
t ≈ 0.91 s
Therefore, the ball was thrown approximately 0.91 seconds before it passed the window.
D. To find how long after the baseball passed the window it reaches the street again, we can use the equation for vertical motion:
s = ut + (1/2)at^2
Where:
s = displacement (28.1 m, the initial altitude above the street)
u = final velocity (-13.7 m/s, as it falls downward)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (what we need to find)
By substituting the values into the equation, we can solve for t:
28.1 = -13.7t + (1/2)(-9.8)t^2
0 = -4.9t^2 - 13.7t + 28.1
Using the quadratic formula:
t ≈ 1.77 s
Therefore, the baseball reaches the street again approximately 1.77 seconds after it passed the window.
To solve these questions, we need to use kinematic equations and principles of motion. Let's break down each question and explain how to find the answer.
A. To find the initial speed of the baseball, we can use the equation for vertical motion:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. In this case, we are given the final velocity (13.7 m/s) and the vertical acceleration (-9.8 m/s^2). We need to find the initial velocity (vi).
Since the baseball is passing upward, the final velocity will be negative. Rearranging the equation, we have:
vi = vf - at
Substituting the given values, we get:
vi = 13.7 m/s - (-9.8 m/s^2) * t
B. To find the altitude reached by the baseball, we can use the equation for vertical displacement:
Δy = vi * t + (1/2) * a * t^2
where Δy is the displacement, vi is the initial velocity (which we found in question A), a is the acceleration due to gravity, and t is the time.
In this case, we are interested in the altitude reached by the baseball, which means Δy is the height above its starting position. Therefore, Δy is equal to the given height of the window (28.1 m) plus the initial height of the baseball from the street.
By rearranging the equation, we get:
28.1 m = vi * t + (1/2) * (-9.8 m/s^2) * t^2
C. To find the time before the ball passed the window at which it was thrown, we can rearrange the equation we used in question B:
28.1 m - (1/2) * (-9.8 m/s^2) * t^2 = vi * t
Substituting the value we found for vi in question A, the equation becomes:
28.1 m - (1/2) * (-9.8 m/s^2) * t^2 = (13.7 m/s - (-9.8 m/s^2) * t) * t
D. To find the time after the baseball passed the window at which it reaches the street again, we need to find the total time of flight first. We can use the fact that the time of flight is twice the time it takes for the ball to reach its maximum height.
To find the time it takes for the ball to reach its maximum height, we can use the equation:
vf = vi + at
At maximum height, the final velocity (vf) will be zero. Rearranging the equation, we have:
0 = vi + (-9.8 m/s^2) * t_max
Solving for t_max, we get:
t_max = vi / 9.8 m/s^2
Since the total time of flight is twice the time to reach the maximum height, we have:
t_total = 2 * t_max
Once we have the total time of flight, we can find the time it takes for the baseball to reach the street again by subtracting the time at which the ball passed the window from the total time of flight:
t_after_window = t_total - t
where t is the time found in question C.