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
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