A baseball is hit by Aaron Judge, a New York Yankees baseball team member, at an amazing velocity of 110 miles/hour at an angle of 35 degrees from the horizontal from home plate in Yankee Stadium in New York City. The baseball was initially launched from an elevation of 4 feet above ground. The distance from home plate to the outfield wall is 408 feet at center field. The height of the outfield wall is approximately 8 feet (It varies slightly across the outfield). You should first draw a two-dimensional picture of the baseball’s trajectory (x, y coordinate system) to assist in developing answers to the questions below. Note: Convert all variables to the metric (mks) system before performing your calculations.
b)What is the maximum height reached by the ball?
c)What is the total time of flight?
No I will not use miles and feet
49.2 meters/sec = 110 mph
u = horizontal speed = 49.2 cos 35 = 40.3 m/s
that is constant
so
x = 40.3 t
vertical problem:
v = Vi - 9.81 t
Vi = 49.2 sin 35 deg = 28.2 m/s
so
v = 28.2 -9.81 t
at top, v = 0
so t = 2.87 seconds to top
h = top = Hi + Vi t - 4.9 t^2
Hi = 4 feet = 1.22 meters
h = 1.22 + 28.2(2.87)-4.9(2.87)^2
h = 1.22 + 80.9 - 40.4
= 41.7 meters at top
when does it reach 8 feet?
2.44 = 1.22 + 28.2 t -4.9 t^2
solve quadratic for t
use range = u*t to see if it cleared the wall
if it did solve for t at h = 0
To answer these questions, we first need to analyze the motion of the baseball. We can break down its trajectory into two components: horizontal and vertical.
To begin, let's convert the initial velocity of the baseball from miles per hour to meters per second, as the metric system is preferred for calculations.
1 mile = 1609.34 meters
1 hour = 3600 seconds
So, the initial velocity of the baseball is:
110 miles/hour * (1609.34 meters/1 mile) * (1 hour/3600 seconds) = 49.17 meters/second.
Now, let's define the variables we'll be using:
- vi: initial velocity of the baseball (49.17 m/s)
- θ: angle from the horizontal (35 degrees)
- y0: initial elevation of the baseball (4 feet = 1.2192 meters)
- g: acceleration due to gravity (9.8 m/s^2)
b) To find the maximum height reached by the ball, we need to determine the vertical component of the initial velocity (viy). This can be calculated as follows:
viy = vi * sin(θ)
viy = 49.17 m/s * sin(35 degrees) = 28.33 m/s
Next, we can use the kinematic equation for vertical motion to find the maximum height (ymax) reached by the ball:
ymax = y0 + (viy^2 / (2 * g))
ymax = 1.2192 m + (28.33 m/s)^2 / (2 * 9.8 m/s^2)
ymax = 1.2192 m + 403.8949 m^2/s^2 / (19.6 m/s^2)
ymax ≈ 1.2192 m + 20.6285 m ≈ 21.85 m
Therefore, the maximum height reached by the ball is approximately 21.85 meters.
c) To find the total time of flight, we need to determine the time it takes for the ball to reach its maximum height (t1) and then double it to account for the descent back to the ground.
The time to reach maximum height can be calculated using the vertical component of the initial velocity (viy) and the acceleration due to gravity (g):
t1 = viy / g
t1 = 28.33 m/s / 9.8 m/s^2 ≈ 2.89 s
The total time of flight is then calculated by doubling t1:
Total time of flight = 2 * t1 ≈ 2 * 2.89 s ≈ 5.78 s
Therefore, the total time of flight for the baseball is approximately 5.78 seconds.