The motion of a ball scooped by a field hockey player can be modeled by h= -16t + 40t, where t is the time in seconds and h is the height of the ball. Will the ball ever reach 22 feet?
I don't understand how to solve this. Help?
the 1st term should be -16t^2 , it reflects gravity's action during freefall
40 is the initial upward velocity of the ball
plug in 22 for h , and solve for t
So
22= -16t^2 + 40t
Subtract 22
0= -16t^2 + 40t -22
Now what?
you've heard of the quadratic formula?
h = -16 t^2 + 40 t
parabola, where is vertex?
16 t^2 -40 t = -h
t^2 - 2.5 t = -h/16
t^2 - 2.5 t + 1.25^2 = -h/16 + 1.5625
(t - 1.25)^2 = -(1/16) (h-25)
vertex (top) at t = 1.25 seconds and h = 25 feet
so it passes 22 on the way up and on the way down :)
Okay thank you
To determine if the ball will ever reach a height of 22 feet, we can set up an equation by equating the given height, h, to 22 feet and solve for time, t.
The equation given to model the motion of the ball is h = -16t^2 + 40t.
Substituting h with 22, we get:
22 = -16t^2 + 40t
To solve this equation, we need to rearrange it into a quadratic equation form by moving all terms to one side:
-16t^2 + 40t - 22 = 0
Now, we can solve this quadratic equation using various methods such as factoring, completing the square, or using the quadratic formula.
In this case, the quadratic equation cannot be easily factored, so we'll use the quadratic formula, which states that for an equation in the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation -16t^2 + 40t - 22 = 0, the coefficients are:
a = -16
b = 40
c = -22
Plugging these values into the quadratic formula, we have:
t = (-40 ± √(40^2 - 4(-16)(-22))) / (2(-16))
Simplifying further:
t = (-40 ± √(1600 - 1408)) / (-32)
t = (-40 ± √192) / (-32)
t = (-40 ± √(16*12)) / (-32)
t = (-40 ± 4√3) / (-32)
Now, we have two possible values for t:
t₁ = (-40 + 4√3) / (-32)
t₂ = (-40 - 4√3) / (-32)
By calculating these values, we find that:
t₁ ≈ 2.8 seconds
t₂ ≈ -0.2 seconds
The negative value for t₂ represents an invalid solution because time cannot be negative in this context.
Thus, the ball will reach a height of 22 feet (approximately) at around 2.8 seconds.