Question

A golfer hits a ball from a starting elevation of 2 feet with a velocity of 70 feet per second down to a green with an elevation of −7 feet. The number of seconds t it takes the ball to hit the green can be represented by the equation −16t2 + 70t + 2 = −7. How long does it take the ball to land on the green?

Answer 1

Answer:

4.50 s

Step-by-step explanation:

The motion of the ball is represented by the following equation:

$h(t) = 2+70t-16t^2$ (1)

where

h(t) is the elevation at time t

2 ft is the initial elevation at t = 0

+70 ft/s is the initial vertical velocity

$-32 ft/s^2$ is the acceleration due to gravity

The green is located at an elevation of -7 feet, so the ball lands on the green when

h(t) = -7

Substituting into (1)

$-7=2+70t-16t^2$

And re-arranging we get

$16t^2-70t-9=0$

This is a second-order equation in the form

$at^2+bt+c=0$

which has solutions

$t_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (2)

Here we have:

a = 16

b = -70

c = -9

Substituting into (2) we find the solutions:

$t_{1,2}=\frac{70\pm \sqrt{(-70)^2-4(16)(-9)}}{2(16)}$

The two solutions are:

t = 4.50 s

t = -0.125 s

The second solution is negative, and since negative time has no physical meaning, the only correct solution is

t = 4.50 s