If a ball is thrown vertically upward from a height of 56ft. above ground with an initial velocity of 40ft. per second, then the height of the ball above ground t seconds after it is thrown is given by f(t)=-16t^2 + 40t +56. How many seconds will elapse after the ball is thrown before it hits the ground?
I need step by step explaination please. I know the answer is 3.5 seconds.
Well, you have a function. If f(t) = the height, when does f=0?
In other words, just solve the dang quadratic!
-16t^2 + 40t +56 = 0
t = -1 and 3.5
See whether you can pick the proper one.
okay, so I solved as follows:
-16t+40t+56=0
(4t+4)(-4t+14)
4t+4=0 and -4t+14=0
4t=4 and -4t=-14
4t=4/4 and -4t= -14/-4
= t=1 and t = 7/2
Now, I am lost at what to do from here.
duh, I got it. you divide 7/2 to equal 3.5
To find out how many seconds will elapse before the ball hits the ground, we need to determine when the height of the ball above the ground is zero.
According to the given equation: f(t) = -16t^2 + 40t + 56
Setting f(t) to zero, we have: -16t^2 + 40t + 56 = 0
To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 40, and c = 56. Plugging these values into the formula, we have:
t = (-40 ± √(40^2 - 4(-16)(56))) / (2(-16))
Simplifying further, we have:
t = (-40 ± √(1600 + 3584)) / (-32)
t = (-40 ± √(5184)) / (-32)
t = (-40 ± 72) / (-32)
Now, we have two possible values for t: (-40 + 72) / (-32) and (-40 - 72) / (-32)
Calculating the two values, we get:
t = 32 / (-32) = -1
t = -112 / (-32) = 3.5
Since time cannot be negative in this context, we discard the negative value. Therefore, the ball will hit the ground after approximately 3.5 seconds.