Answer to Question #259188 in Calculus for zetadddd

Question

An object in free fall from

50 m high, so the height is

when t is h(t) = 50 – 4.9t2

.

What is the speed at

t = 2?

2. Determine the equation of the tangent line

on the curve y = x3 at the point (2.8)

Answer 1

$c\displaystyle \text{The speed is given by the derivative of h(t)}\\ h(t) = -4.9t^2 + 50\\ h'(t) = -9.4t\\ \text{At t = 2,$ h'(t)$ = -18.8}\\ \text{For the 2nd question, the equation of the tangent is given by $y= mx+c$}\\ \text{where m represents the gradient and c the intercept, we find m by computing the }\\ \text{the derivative}\\ y'(x) =3x^2, \text{at x = 2.8, $y = 3(2.8)^2 =23.52$}\\ \text{Next, we compute the intercept c, at x = 2.8}\\ 2.8^3 = 23.52(2.8) + c\\ \implies c = -43.904\\ \text{Thus, the equation of the tangent is given by}\\ y = 23.52x -43.904$