The vertical acceleration of a surveillance drone, a(t)a(t)a(t) in m s−2\text{m s}^{-2}m s−2, is modeled by the function H′′(t)=12t−4t2H''(t) = 12t - \dfrac{4}{t^2}H′′(t)=12t−t24 for t>0.5t > 0.5t>0.5, where t t\,t is the time in seconds after launch and H(t)H(t)H(t) is the height in metres.
At the instant t=1t = 1t=1, the height and vertical velocity of the drone are such that the tangent to the graph of H H\,H against t t\,t has the equation H=8t−5H = 8t - 5H=8t−5.
Find an equation of the normal to the graph of H H\,H at the point where t=1t = 1t=1.
Find H(t)H(t)H(t), writing your answer in simplest form.
