Question

A ball is thrown in the air. Its height at any time \(t\) is given by \(h = 3+14t-5t^2\), then the maximum height it can reach

• A. \(12.9\)
• B. \(12.8\)
• C. \(12.7\)
• D. \(12.6\)

Hint:

Differentiate h, set the derivative to zero and put that time back into h.

Answer 1

The Correct Option is B

Step 1: Complete the square:
$h = -5t^2 + 14t + 3 = -5\left(t^2 - \tfrac{14}{5}t\right) + 3$.

Step 2: Rewrite:
$h = -5\left(t - \tfrac{7}{5}\right)^2 + 5\cdot\tfrac{49}{25} + 3 = -5\left(t - 1.4\right)^2 + 9.8 + 3$.

Step 3: Read the maximum:
The square term is never positive, so the largest value is $9.8 + 3 = 12.8$, reached at $t = 1.4$. This is option (B).

Final Answer:
The maximum height is 12.8. \[ \boxed{\text{(B) }12.8} \]