A ball is thrown vertically upwards. Its height h (in metres) above the ground after t seconds is \(h = 40t - 5t^{2}\). The maximum height reached by the ball is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 80 m
The height is greatest when its rate of change (the velocity) is zero:
\( \displaystyle \frac{dh}{dt} = 40 - 10t = 0 \)
\( t = 4 \)
Since \(\frac{d^{2}h}{dt^{2}} = -10 \lt 0\), this is a maximum. The height then is
\( h = 40 \times 4 - 5 \times 16 \)
\( = 160 - 80 = 80 \) m
Why the other options are incorrect
Option (1) 40 m — Incorrect. This is the starting speed 40 m/s, not a height.
Option (2) 160 m — Incorrect. This is only the term 40t at t = 4; the term 5t2 = 80 has not been subtracted.
Option (3) 75 m — Incorrect. This is the height at t = 3 s, before the ball reaches the top.