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?
a. yours, including conversions.
b. max height is when vy is zero
vy=vyi-g*t
vyi=initial velociyt* sin4
vy=zero, solve for time t.
then
hmax=vyi*t-4.9t^2
c. time of flight
hf=hi+viy*t-4.9t^2
hf=8ft converted to m
hi=half that in m.
folve for time t.
oops, angle is 35, so sin35.
Thank you for the equation listed above! I am trying to solve it but I believe I am not doing a great job on it. Would you go step by step with me please? I'd appreciate it!
post it step by step, I will check it.
The equation I start off with is vy=vyi-g*t?
So it would be 0=4*sin35?
To calculate the maximum height reached by the ball, we can use the kinematic equation for vertical motion. The equation is:
y = y0 + v0y*t - (1/2)*g*t^2
Where:
y is the height (vertical displacement) of the ball at any given time t
y0 is the initial height (elevation) of the ball above the ground, which is 4 feet converted to meters (1.22 meters)
v0y is the initial vertical component of the ball's velocity (110 miles/hour converted to meters/second)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time
To find the maximum height, we need to find the time at which the ball reaches its highest point. At the highest point, the vertical velocity of the ball becomes zero, so v0y - g*t = 0.
Solving for t:
t = v0y / g
Now, to calculate the maximum height (y), we substitute this time back into the equation for y:
y = y0 + v0y*t - (1/2)*g*t^2
To convert the velocity from miles/hour to meters/second, we multiply it by the conversion factor (1.60934 km/hour = 1 mile/hour), and divide by 3.6 (1 hour = 3600 seconds).
So, the calculations are as follows:
Convert 110 miles/hour to meters/second:
110 * 1.60934 / 3.6 = 49.03 m/s
Substitute the values into the equation:
t = 49.03 / 9.8 = 5 seconds
y = 1.22 + (49.03 * 5) - (0.5 * 9.8 * (5^2))
y = 1.22 + 245.15 - (0.5 * 9.8 * 25)
y = 1.22 + 245.15 - 122.5
y = 123.87 meters
Therefore, the maximum height reached by the ball is approximately 123.87 meters.
To calculate the total time of flight, we can use the formula for the time of flight (T) of a projectile:
T = 2 * t
Since we have already found that the time it takes for the ball to reach its highest point is 5 seconds, substituting this into the formula gives us:
T = 2 * 5 = 10 seconds
Therefore, the total time of flight of the ball is 10 seconds.