Question

A half ball is hit with an initial vertical veolicty of 68 ft/sec. The function h=-16t^2+68t models the height h (in feet) of the golf ball at time t (in seconds). Does the golf ball reach a height of 70ft? If so what time? What is the maximum height the ball reaches?

Answer 1

Answer:

Yes; at 1.75 s; 72.25 ft

Step-by-step explanation:

h = -16t² +68t

a = -16; b = 68; c = 0

The vertex h of a parabola is at

h  = -b/(2a) and the maximum height is at

y = f(h)

1. Time to maximum height

$t = -\dfrac{b}{2a} = -\dfrac{68}{2\times(-16)} = \dfrac{68}{32} = \textbf{2.125 s}$

2. Maximum height

$\text{Height} = f(2.125) = -16(2.125)^{2} + 68(2.125) = -72.25 + 144.5 = \textbf{72.25 ft}$

3. Does the golf ball reach 70 ft?

Yes, it passes 70 ft on the way to its maximum height of 72.25 ft.

4. Time to 70 ft

      -16t² + 68t = 70

-16t² + 68t - 70 = 0

   8t² -34t + 35 = 0

  ( 4t -7) (2t - 5) = 0

4t - 7 = 0          2t - 5 = 0

    4t = 7                2t = 5

      t = 1.75 s          t = 2.5 s

The golf ball reaches 70 ft at 1.75 s on the way up and 2.5 s on the way down.

The diagram below shows the path of your parabola.