s(t) = –16t^2 + 128t + 30
can do this with or without calculus
using calculus
ds/dt = -32 t + 128
= 0 at max height
32 t = 128
t = 4 seconds
for max height
s(4) = -16(16) +128(4) + 30
s(4) = 286
without calculus find vertex of parabola
16 t^2 -128 t = -s+30
complete square
t^2 - 8t = -(1/16)s + (30/16)
t^2 - 8 t + 16 = -(1/16)s + 15/8 + 16
(t-4)^2 = etc
so vertex at t = 4
If a projectile is launched from a platform 30 feet high with an initial velocity of 128 feet per second, then the height of the projectile at t seconds is given by s(t) = –16t2 + 128t + 30 feet.
(a) At what time does the projectile attain its maximum height? Show some work.
(b) What is the maximum height attained by the projectile? Show some work.
1 answer
1 answer