Question

the empire state building is 1250 feet tall. IF an object is thrown upward from the top of the building at an initial velocity of 38 feet per second, its height s seconds after it is thrown is given by the function h(s) = -16s^2 + 38s + 1250. round to the nearest hundreth

Answer 1

Answer:

Time for the object to get h(max)    s = 1.1875 sec

h (max)  = 1272.57 feet

Down time for the object to hit the ground   =  4.25 sec

Step-by-step explanation:

The relation

h(s)  =  -  16*s²  +  38* s  + 1250    (1)

Is equivalent to the equation for vertical shot

Δh = V(i)*t  -  1/2g*t²  (in this case we don´t have independent term since the shot is from ground level. We can see in (1), the independent term is 1250 feet ( the height of the empire state building), the starting point of the movement.

The description of the movement is:

V(s)  =  V(i)  - g*s     ⇒  V(s) = 38 - 32*s

At h(max)    V(s)  =  0      38/32  = s

So the maximum height is at  s = t = 1.1875 sec

The time for the object to pass for starting point is the same

t  =  1.1875 sec

h(max) is

h(max)  = - 16* (1.1875)² +  38 (1.1875) + 1250

h(max)  =  -  22,56  +  45.13  +  1250

h(max)  =  1272.57 feet

Time for the object to hit the ground is

h(s)  =  - 1250 feet

-1250  =  - 16 s² + 38*s  + 1250

-16s² +  38s  =  0

s ( -16s + 38 )  =  0

First solution for that second degree equation  is x = 0 which we dismiss

then  

( -16s + 38 )  =  0    ⇒ 16s   =  38     s  =  38/16

s =  2.375 sec    and  we have to add time between h (max) and to get to starting point  ( 1. 1875 sec)

total time is  = 2.375 + 1.875

Total time  =  4.25 sec