well, if you are not allowed to use calculus you must complete the square to find the vertex of that parabola.
16 t^2 - 48 t -20 = -h
t^2 - 3 t - 1.25 = -h/16
t^2 - 3 t = -h/16 + 1.25
t^2 - 3 t + 9/4 = -h/16 + 3.5
(t - 1.5)^2 = - 1/16 ( h - 56)
so at peak t = 1.5 and h = 56
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check with calculus
when h is max, dh/dt = 0
0 = -32 t + 48
so t = 48/32 = 1.5 sure enough
then h = -16(2.25) + 48(1.5) + 20
= 56 ok
A projectile is launched from a platform 20 feet high with an initial velocity of 48 feet per second, The height h of the projectile at t seconds after launch is given by h = –16t^2 + 48t + 20 feet.
(a) How many seconds after launch does the projectile attain maximum height? (b) What is the maximum height
1 answer
1 answer