A ball is thrown into the air with an upward velocity of 40 feet per second. Its height, h, in feet after t seconds is given by the function h(t)=-16t^2+40t+10.
a. What is the ball’s maximum height? 35Ft
b. When the ball hits the ground, how many seconds have passed? Show your work
16 t^2 - 40 t - 10 = -h
divide by 16 first to get 1 as coef of t^2
t^2 - 2.5 t - .625 = -h/16
t^2 - 2.5 t = -h/16 + .625
half of 2.5 then square
get 1.5625, add to both sides
t^2 - 2.5 t + 1.5625 =-h/16 + 1.5625
(t-1.25)^2 = -(1/16)(h-25)
vertex (top of the parabola) at
h = 25
t = 1.25 seconds
now solve for t when h = 0 .
Use the + answer
I assume you do not do calculus so find the vertex of that parabola. That is just like the last one we did :)
I will race you.
The vertex is (5/4,35)
Beat ya :)
anyway part 2
16 t^2 - 40 t - 10 = 0
quadratic equation
a = 16
b = -40
c = -10
we agree about t, but not about h
CHECK
16 t^2 - 40 t - 10 = -h
16(1.25)^2 - 40(1.25) - 10 = ?
25 - 50 -10
-35 = -h , you are right
I forgot to add 10 on the right
t = 2.73 for ground time
https://www.mathsisfun.com/quadratic-equation-solver.html
To find the maximum height of the ball, we need to determine the vertex of the quadratic function h(t) = -16t^2 + 40t + 10. The vertex of a parabola is given by the formula (-b/2a, f(-b/2a)), where a, b, and c are the coefficients of the quadratic equation.
In this case, a = -16, b = 40, and c = 10.
Using the formula, the x-coordinate of the vertex is given by:
x = -b / (2a)
= -40 / (2 * -16)
= -40 / -32
= 1.25
To find the maximum height, we substitute t = 1.25 into the h(t) equation:
h(1.25) = -16(1.25)^2 + 40(1.25) + 10
= -16(1.5625) + 50 + 10
= -25 + 60
= 35
Therefore, the ball's maximum height is 35 feet.
To determine how many seconds have passed when the ball hits the ground, we need to find the value of t when h(t) equals zero since h(t) represents the height of the ball.
Setting h(t) = 0:
-16t^2 + 40t + 10 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. However, in this case, factoring would not be possible, so we will use the quadratic formula.
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
Using the formula, the values are:
a = -16, b = 40, c = 10
t = (-40 ± √(40^2 - 4(-16)(10))) / (2(-16))
= (-40 ± √(1600 + 640)) / (-32)
= (-40 ± √2240) / (-32)
Simplifying further:
t ≈ (-40 ± 47.35) / (-32)
Now, we have two possible solutions for t:
t1 ≈ (-40 + 47.35) / (-32)
≈ 7.35 / (-32)
≈ -0.23
t2 ≈ (-40 - 47.35) / (-32)
≈ -87.35 / (-32)
≈ 2.73
Since time cannot be negative in this context, we discard t1 as an extraneous solution.
Therefore, approximately 2.73 seconds have passed when the ball hits the ground.