Amir pitches a baseball at an initial height of 6 feet with a velocity of 73 feet per second. This can be represented by the function H(t) = −16t2 + 73t + 6. If the batter misses, about how long does it take the ball to hit the ground?
4.64 seconds
2.94 seconds
2.28 seconds
0.08 seconds
The ball will hit the ground when H(t) = 0. So we need to solve the equation:
0 = -16t^2 + 73t + 6
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
We have a=-16, b=73, and c=6
t = (-73 ± sqrt(73^2 - 4(-16)(6))) / 2(-16)
t = (-73 ± sqrt(5329)) / -32
t = (-73 ± 73) / -32
t = 0.08 or 4.64
Since the ball was initially pitched at a height of 6 feet, it is the first solution we want, not the second. Therefore it takes about 0.08 seconds for the ball to hit the ground.
The answer is: 0.08 seconds
To find the time it takes for the ball to hit the ground, we need to determine when the height of the ball, H(t), is equal to zero.
Given the function H(t) = −16t^2 + 73t + 6, we can set it equal to zero and solve for t:
−16t^2 + 73t + 6 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 73, and c = 6.
Plugging in these values into the quadratic formula, we get:
t = (-(73) ± √((73)^2 - 4(-16)(6))) / (2(-16))
Simplifying further:
t = (-73 ± √(5329 + 384)) / (-32)
t = (-73 ± √(5713)) / (-32)
t ≈ 4.64 seconds or t ≈ 0.08 seconds
Since the ball is initially thrown from a height of 6 feet, and the height of the ball at t = 0 is 6 feet, we can conclude that the ball hits the ground at t ≈ 4.64 seconds.
Therefore, the correct answer is 4.64 seconds.