h=-16t2+20t+24 gives the height h of the rock after t seconds. how many seconds does it take for the rock to hit the water
24+20=16t
44=16t
44/16=16/16
2.75 seconds
when it hits the water, h = 0
0 = -16t^2 + 20t + 24
4t^2 - 5t - 6 = 0
(t-2)(4t + 3) = 0
t = 2 or t = -3/4, the last answer is inadmissible
It will take 2 seconds
I have no idea what the last part of your post is.
Is that supposed to be a solution ?
If so, it makes no sense at all.
H=-16t2+24t+16. Find the height after 1 seconds.
To find the number of seconds it takes for the rock to hit the water, we need to find the value of t when the height h is equal to zero. In this case, the equation is h = -16t^2 + 20t + 24.
Setting h to zero, we get:
0 = -16t^2 + 20t + 24
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = 20, and c = 24.
Substituting these values into the quadratic formula:
t = (-(20) ± sqrt((20)^2 - 4(-16)(24))) / (2(-16))
Simplifying further:
t = (-20 ± sqrt(400 + 1536)) / (-32)
t = (-20 ± sqrt(1936)) / (-32)
t = (-20 ± 44) / (-32)
This gives two possible values for t: (-20 + 44) / (-32) and (-20 - 44) / (-32)
Simplifying:
t = 24 / -32 = -0.75
t = -64 / -32 = 2
Since time cannot be negative in this context, we discard the value of -0.75. Therefore, the rock hits the water after 2 seconds.