A batter hit a baseball at a height 3 feet off the ground, with an initial vertical velocity of 64 feet per second. The function that models the height of the ball as a function of time is ℎ(푡) = −16푡 2 +64푡 +3. a. If the ball was caught after 4 seconds, what was the height of the ball when it was caught? b. What was the maximum height reached by the baseball?

1 answer

To analyze the height of the baseball using the function \( h(t) = -16t^2 + 64t + 3 \), we will address both parts of your question.

Part a: Height of the ball when caught

To find the height of the ball when it was caught after 4 seconds, we will evaluate the function \( h(t) \) at \( t = 4 \).

\[ h(4) = -16(4^2) + 64(4) + 3 \] \[ = -16(16) + 64(4) + 3 \] \[ = -256 + 256 + 3 \] \[ = 0 + 3 = 3 \]

So, the height of the ball when it was caught is 3 feet.

Part b: Maximum height reached by the baseball

To find the maximum height of the baseball, we need to determine the vertex of the quadratic function. The height function \( h(t) \) is a parabola that opens downward (since the coefficient of \( t^2 \) is negative). The time \( t \) at which the maximum height occurs can be found using the formula:

\[ t = -\frac{b}{2a} \]

Here, \( a = -16 \) and \( b = 64 \).

\[ t = -\frac{64}{2(-16)} = \frac{64}{32} = 2 \]

Now we can find the maximum height by evaluating \( h(t) \) at \( t = 2 \).

\[ h(2) = -16(2^2) + 64(2) + 3 \] \[ = -16(4) + 64(2) + 3 \] \[ = -64 + 128 + 3 \] \[ = 64 + 3 = 67 \]

Thus, the maximum height reached by the baseball is 67 feet.

Summary of Results:

  • a. Height when caught: 3 feet
  • b. Maximum height reached: 67 feet