h=-16t^2+90t+3?

Question

How long will it take for the object to hit the ground?

Answer 1

h = -16t^2 + 90t + 3
h = 0

16t^2 - 90t - 3 = 0
t = [90+-sqrt(8292)]/32
= 5.658s

The object will hit the ground after about 5.66 seconds

Answer 2

the object will hit the ground when h=0.
-16t^2 + 90t + 3 = 0
16t^2 - 90t - 3 = 0
t = [ 90 +or- sqrt(8100 + 4(16)3 ] / 32
= [ 90 +or- sqrt(8292) ] / 32
= [ 90 + or- 91.06 ] / 32
take t >0
t = 5.66 sec

Answer 3

Well, since this is a function for height, the ground would be defined to be the point where height is equal to zero. Therefore, we'll just set this equation equal to zero and solve for time:

0 = -16t^3 + 90t + 3

There's no better way to do this than the quadratic equation, so let's use it (a = -16, b = 90, c = 3)

t = [ -b ± √(b² - 4ac) ] / (2a)
t = [ -90 ± √(90² - 4(-16)(3)) ] / (2(-16))
t = [ -90 ± √8292 ] / (-32)
t = [ -90 ± 91.06 ] / (-32)
t = -181.06 / -32 ... OR... t = 1.06 / -32
t = 5.6581 seconds ... OR ... t = -0.0331 seconds

Since we can't have a negative value for time, we discard the second value, and we get our final answer (rounded to 3 decimal places) of t = 5.858 seconds.

Answer 4

when the object hits the ground then h=0 so:

16t^2 - 90t - 3 = 0

use x = [-b +- sqrt(b^2 -4ac) ]/ 2a

a= 16
b= -90
c= -3

x= 181.0604/32 = 5.658 sec
or x = -1.06 sec (not realistic)