please help with this math function problem?

Question

The function h described by h(t) = - 16t2 + 33.1t + 124.26 gives the height of a ball thrown upward with a speed of 33.1 feet per second from a 124.26 ft high window t seconds after it is thrown until it hits the ground. Find the height of the ball 1.2 seconds after it is thrown.

Answer 1

h(t) = - 16t2 + 33.1t + 124.26
h(1.2) = -16(1.2)^2 +33.1(1.2) +124.26

= -23.04 +39.72 +124.26
=140.94

Answer 2

I don't understand where you have the problem. You have the function. You have the values of the variables. And you have them set up in an equation.

Distance (d) measured from a starting point is calculated according to the following formula.
d=(v0)t + (1/2)at^2
v0=33.1 (initial velocity... given)
t=1.2 (the time over which all this stuff takes place...given)
a= -32 (acceleration due to gravity... negative because it's decreasing the height... ;a constant)

Substituting
d=(1.2)(33.1) + (1/2)(-32)(1.2)^2 NOTE: (1/2)(-32) is where the -16 comes from.

The reason you have to add 124.26 to it because you started that far above the ground.

h = d + 124.26

You can use a calculator as well as I can... probably better.

Answer 3

h = -16(1.2)² + 33.1(1.2) + 124.26
h = -23.04 +39.72 + 124.26
h = 140.94 ft

Answer 4

h(t)=-16t^2+33.1t+124.26
just replace t with 1.2
h(1.2)=-16(1.2)(1.2)+33.1(1.2)+124.26
h(1.2)=-23.04+39.72+124.26
h(1.2)=140.94 ft