A batter hits a ball 3 feet from home plate along the path x=69t, y=3+40t-16t^2. How long is the ball in flight and how far does it travel?
the ball is in flight till y=0:
t=2.57
after 2.57 seconds, x=177.53 ft
note that the problem should have said the ball was hit 3 ft above</it> home plate.
To find the duration of the ball's flight, we need to determine the time at which it lands on the ground, meaning when its y-coordinate equals zero. We can set the equation for y equal to zero and solve for t:
y = 3 + 40t - 16t^2
0 = 3 + 40t - 16t^2
We have a quadratic equation (16t^2 - 40t + 3 = 0), which can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 16, b = -40, and c = 3. Plugging these values into the formula, we get:
t = (-(-40) ± √((-40)^2 - 4*16*3))/(2*16)
t = (40 ± √(1600 - 192))/(32)
Now, we simplify the expression further:
t = (40 ± √1408)/(32)
t ≈ (40 ± 37.5)/(32)
Now, we have two possible solutions for t:
t1 ≈ (40 + 37.5)/(32) ≈ 2.9 seconds
t2 ≈ (40 - 37.5)/(32) ≈ 0.1 seconds
Since we are considering a time when the ball is in flight, we discard the t2 value of approximately 0.1 seconds. Therefore, the ball is in flight for approximately 2.9 seconds.
To determine the distance the ball travels, we need to find the horizontal distance traveled, which is the x-coordinate when the ball lands on the ground. We substitute the value of t1 into the equation for x:
x = 69t
x = 69(2.9)
x ≈ 200.1 feet
Therefore, the ball is in flight for approximately 2.9 seconds and travels approximately 200.1 feet.