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balandron [24]
3 years ago
7

A football player kicks a ball with an initial upward velocity of 47 ft./s. The initial height of the ball is 3 feet. The functi

on h(t)=-16t^2+vt+h0 models the height (in feet) of the ball, where V is the initial upward velocity and h0 is the initial height. If no one catches the ball, how long will it be in the air?
Mathematics
1 answer:
Rashid [163]3 years ago
6 0

Answer: 3 seconds

We are given the following function:

h(t)=-16{t}^{2}+V.t+h(o)    (1)

Where:

t is the time the ball is in the air

V=47ft/s the initial upward velocity

h(o)=3ft the initial height of the ball

h(t) is the final height of the ball. If no one catches it, this will be <u>zero</u>

So, equation (1) changes to:

-16{t}^{2}+V.t+h(o)=0    (2)

Substituting the known values:

-16{t}^{2}+(47ft/s).t+3ft=0    (3)

This is a quadratic equation in the form a{t}^{2}+b.t+c=0. In order to find t we can use the quadratic formula for the roots:

t=\frac{-b\pm\sqrt{-4ac}}{2a}    (4)

Where a=-16, b=47 and  c=3

Substituting this values in (4):

t=\frac{-47\pm\sqrt{-4(-16)(3)}}{2(-16)}    (5)

t=\frac{-47\pm\sqrt{2401}}{-32}    (6)

t=\frac{-47\pm\49}{-32}    (7)

For t1:

t1=\frac{-47+49}{-32}=-0.0625 s      (8)>>>> This result does not work for us because is negative

For t2:

t2=\frac{-47-49}{-32}=3 s   (9)>>>This is the result

Therefore:

<h2> If no one catches the ball, it will be 3 s in the air</h2>
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