An object is lauched from a platform. The object’s height above the ground after t seconds is given by the function: h(t) = −16t2 + 20t + 50.

Question

How high was the platform?

(b) What is the maximum height of the object?

(c) How high is the object at 1 second?

(d) How long was the object in the air?

Answer 1

50 foot high platform

max height is when h'(t) = 0 = -32t+20

t= 20/32 = 5/8 second

h(5/8) = -16t^2 +20t +50 = -16(5/8)^2+20(5/8)+50

= -25/4 +25/2+50

= 50+25/4= 50+6 1/4

= 56 1/4 feet high = maximum height

after 1 second the height is

h(1) = -16(1)^2 + 20(1) +50

=-16+20+50

= 54 feet

the object was in the air for 2.5 seconds