Question

67. Maximizing the Height of an Object If an object is thrown upward with an initial velocity of 32 ft/second, 32 , ft slash , second , comma then its height after t seconds is given by h(t)=32t−16t2. h , open t close , equals 32 t minus 16 , t squared , . (a) Find the maximum height attained by the object. (b) Find the number of seconds it takes the object to hit the ground.

Answer 1

(a)Therefore the maximum height of the object is 16 feet.

(b)Therefore the object hits the ground after 2 seconds.

Explanation:

(a)

Given function of height is

h(t)= 32t-16t²

where t is time in second

[The unit of h(t) is not given, so consider the unit of h(t) in feet]

We know that,

If a function y = ax²+bx+c.

The maximum value of the y is at
`x=-(b)/(2a)`

Here a= -16 and b =32.

The object attains its maximum height when
`t=-(32)/(2.(-16)) =1 \ s`

Its means after 1 s , the object attains its maximum height.

Therefore the maximum height of the object is

h(1) = 32.1-16(1)² [putting t= 1]

=32-16

=16 feet

(b)

When the object teaches the ground then height of the object will be zero.

i.e h(t) = 0

Therefore,

32t-16t² =0

⇒16t(2-t)=0

⇒t=0 or, t=2

Therefore the object hits the ground after 2 seconds.