A ball is thrown into the air. The height h, in feet, of the ball can be modeled by the equation h= -16t^2+20t+6, where t is the time, in seconds, the ball is in the air. When will the ball hit the ground?
first should I use x= -b/2a?
x = -b/2a will give you the vertex. That will tell you how high the ball went if you plug it in and evaluate h.
To find the time when the ball hits the ground, you need to solve the equation h = 0, since the height will be zero at that time. The given equation is h = -16t^2 + 20t + 6.
Using the quadratic formula, x = (-b ± sqrt(b^2 - 4ac))/(2a), where a, b, and c are the coefficients of the quadratic equation, we can find the values of t when h = 0.
In this case, a = -16, b = 20, and c = 6. Plugging these values into the quadratic formula, we get:
t = (-20 ± sqrt(20^2 - 4*(-16)*6)) / (2*(-16))
Simplifying further:
t = (-20 ± sqrt(400 + 384)) / (-32)
t = (-20 ± sqrt(784)) / (-32)
t = (-20 ± 28) / (-32)
This gives two possible values for t:
t1 = (-20 + 28) / (-32) = 8/(-32) = -1/4
t2 = (-20 - 28) / (-32) = -48/(-32) = 3/2
Since time cannot be negative in this context, the ball will hit the ground at t = 3/2 or 1.5 seconds.
-2(8t^2 -10t -6) = 0
-2(2t -3) (4t +1) = 0
2t -3 = 0
2t-3 + 3 = 0 + 3
2t = 3
t = 3/2 second
Thank you to all
10 feet
Solve the equation
-16t^2+20t+6 = 0
will give you the times that the ball is on the ground. The negative root accounts for the fact that the ball is 6 feet above ground at t=0.
I get t=3/2 and t=-1/4