WAEC - Further Mathematics (2021 - No. 20)

A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).

Calculate the maximum height reached.

418.5m

56.0m

31.5m

30.0m

S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);

ds/dt = 12 + 5t - 3t\(^2\)

At max height ds/dt = 0

i.e 12 + 5t - 3t\(^2\)

(3t + 4)(t -3) = 0;

t = -4/3 or 3

Hmax = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)

= 36 + 45/2 - 27

= 31.5m