Question

an object is propelled upward from the top of a building with an initial velocity of 68ft/sec. the height h (in feet) of the object after t sec is given by
h=-16t^2+68t+60
a) what is the initial height of the object?
b)when is the object 102 ft above the ground?
c)when does the object hit the ground?

Answer 1

\(\large\color{black}{ \displaystyle h(x)=-16t^2+68t+60 }\)

`PART [A]`
The intial height is at 0 seconds (or, at t=0).
Alternatively, the initial height is h(0).

\(\large\color{black}{ \displaystyle h(0)=-16(0)^2+68(0)+60 =\quad? }\)


`PART [B]`
When is the object 102 feet above the ground?
Alternatively, when is the height of the object is 102 feet?
In other words, at what time, or at what value of t do get height 102 or h(t)=102.

\(\large\color{black}{ \displaystyle 102=-16t^2+68t+60 }\)
And solve for t.



`PART [C]`
When does the object hit the ground?
When the object hits the ground, the height is 0, or h(t)=0.

\(\large\color{black}{ \displaystyle 0=-16t^2+68t+60 }\)
And solve for t.
(Exclude negative solutions for t, because time can't be negative)