Question

A rocket is fired straight up from a 60 ft. platform with an initial velocity of 96 ft/sec. The height of the rocket, h(t), is found using the function h(t) = - 16t2 + 96t + 60 where t is the time in seconds.

Find the maximum height.




190 ft


7 ft


204 ft


202 ft

Answer 1

the physics really is just extra info

they just want to know the maximum value for the function h(t)

Answer 2

h(t) = -16t^2 + 96t + 60
v(t) = -32t + 96
at max height , v = 0
-32t + 96=0
32t = 96
t = 96/32 = 3

h(3) = -16(9) + 96(3) + 60 = 204

or

h(t) = -16(t^2 - 6t + 9-9) + 60
= -16( (t-3)^2 - 9) + 60
= -16(t-3)^2 + 144 + 60
= -16(t-3)^2 + 204

this has a vertex of (3,204)
thus a max of 204 when t - 3

Answer 3

yeah the derivative works to figure when the velocity is zero at the maximum point

Answer 4

or show him the derivation for the kinematic equations

Answer 5

so i was right? 204?

Answer 6

you said for algebra though, the last time you just remembered the axis of symmetry thing
t = -B/(2A)
= -96/(2*-16) = 3

the maximum value for that flipped parabola is on that symmetry line, and is the vertex point

h(3)