h(t) = -16t^2 + 48t + 160

Question

The formula h(t) = -16t^2 + 48t + 160 gives the height of an object thrown from a building 160 feet high with an initial speed of 48 ft/sec, where t is measured in seconds. What is the maximum height of the object?

Answer 1

h(t) = -16t^2 + 48t + 160

take the derivative and set =0, as velocity =0 at maximum height

v(t) = h'(t) = -32t + 48 = 0

t = 48/32 = 3/2 = 1.5 seconds to reach maximum height

plug that t value into the original equation to solve for max height

h(3/2) = -16(3/2)^2 + 48(3/2) + 160 = maximum height

h(3/2) = -16(3^2)/2^2 +24(3) + 160

= -4(9) +72 + 160

=-36+72 +160

= 36 +160

= 196 feet