A ball is thrown into the air with an initial upward velocity of 46FT/S its height (h) in feet after T seconds is given by the function H = -16t^2 + 46t + 6. After how many seconds will the ball hit the ground?
A.3
B.4
C.5
D.6
Please I seriously need help I’m super far behind and need to get this done I have no idea what the answer could be I just need help.
it hits the ground when the height is zero, right? So just solve
-16t^2 + 46t + 6 = 0
in the usual ways.
To solve this problem, we need to find the time at which the ball hits the ground. This is equivalent to finding the time when the height of the ball is zero.
Given the height function H = -16t^2 + 46t + 6, we can set the height H equal to zero and solve for t:
0 = -16t^2 + 46t + 6
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = 46, and c = 6. Plugging these values into the quadratic formula, we have:
t = (-46 ± sqrt(46^2 - 4(-16)(6))) / (2(-16))
Now we can simplify and find the solutions for t:
t = (-46 ± sqrt(2116 + 384)) / (-32)
t = (-46 ± sqrt(2500)) / (-32)
t = (-46 ± 50) / (-32)
t = (4 or -6) / (-32)
So, we have two possible values: t = 4 or t = -6. However, since we are dealing with time, it is not possible for t to be negative. Therefore, the ball will hit the ground after 4 seconds.
So, the correct answer is B.4 seconds.