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: Understanding the Concept:
The height is a downward-opening quadratic in \(t\). Its largest value occurs where the rate of change of height, \(dh/dt\), is zero.

Step 2: Finding the time of maximum height:
\[ \frac{dh}{dt} = 14 - 10t = 0 \Rightarrow t = 1.4 \]
The second derivative is \(\dfrac{d^2h}{dt^2} = -10 < 0\), so this is a maximum.

Step 3: Computing the height:
\[ h(1.4) = 3 + 14(1.4) - 5(1.4)^2 = 3 + 19.6 - 9.8 = 12.8 \]
Options (A), (C) and (D) are 12.9, 12.7 and 12.6, which come from arithmetic slips such as using \(t = 1.5\) or \(t = 1.3\).

Step 4: Check:
Using \(t = 1.3\): \(3 + 18.2 - 8.45 = 12.75\). Using \(t = 1.5\): \(3 + 21 - 11.25 = 12.75\). Both are below 12.8, confirming the peak.

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