Question

18. A soccer ball is kicked upward from a height of 1.5 feet with an initial vertical velocity of 35 feet per second. Its

Answer 1

Answer:

Part a) $-16t^{2}+35t+1.5=0$

Part b)

$a=-16\\b=35\\c=1.5$

Part c) The soccer ball will take 2.23 seconds to reach the ground

Step-by-step explanation:

we have

$h(t)=-16t^{2}+35t+1.5$

where

t  is the time the ball has been in the air, in seconds

h(t) is the height, in feet, of the  soccer ball

Part a) Write an equation that could be used to determine the time the ball traveled before it hit the ground

we know that

When the ball hit the ground, the value of h(t) is equal to zero

so

For h(t)=0

$-16t^{2}+35t+1.5=0$

This equation can be used to determine the time the ball traveled before it hit the ground

Part b) Determine the values of a, b, and c

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-16t^{2}+35t+1.5=0$

so

$a=-16\\b=35\\c=1.5$

Part c)  How long will it take for the soccer ball to reach the ground after it was kicked?

Solve the quadratic equation by the formula

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

we have

$a=-16\\b=35\\c=1.5$

substitute the values

$x=\frac{-35(+/-)\sqrt{35^{2}-4(-16)(1.5)}} {2(-16)}$

$x=\frac{-35(+/-)\sqrt{1,321}} {-32}$

$x_1=\frac{-35(+)\sqrt{1,321}} {-32}=-0.04$  ---> cannot be a solution (is negative)

$x_2=\frac{-35(-)\sqrt{1,321}} {-32}=2.23$

therefore

The soccer ball will take 2.23 seconds to reach the ground