Question

A basketball is thrown with an initial upward velocity of 25 feet per second from a height of 8 feet above the ground. The equation h=-16t^2+25t+8 models the height in feet t seconds after it is thrown. After the ball passes its maximum height, it comes down and then goes into the hoop at a height of 10 feet above the ground. About how long after it was thrown does it go into the hoop?

Answer 1

Answer:

A quadratic equation is in the form of $ax^2+bx+c =0$........[1], then the solution for this equation is given by:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

As per the statement:

The equation is given by:

$h=-16t^2+25t+8$

where, h is the height in feet t seconds after it is thrown.

After the ball passes its maximum height, it comes down and then goes into the hoop at a height of 10 feet above the ground.

⇒h = 10 feet

then;

$-16t^2+25t+8=10$

Subtract 10 from both sides we have;

$16t^2-25t+2=0$

On comparing this equation with [1] we have;

a =16 , b =-25 and c =2

then;

$t= \frac{25 \pm \sqrt{(-25)^2-4(16)(2)}}{2(16)}$

⇒$t= \frac{25 \pm \sqrt{625-128}}{32}$

⇒$t= \frac{25 \pm \sqrt{497}}{32}$

as we want the time when it was falling so ,

$t= \frac{25 + \sqrt{497}}{32}$

Simplify:

$t \approx 1.48$ sec

Therefore, 1.48 sec long after it was thrown does it go into the hoop

Answer 2

So we want to know the later time when h=10 so

10=-16t^2+25t+8

16t^2-25t+2=0

Using the quadratic formula for expediency...

t=(25±√497)/32, we want the time when it was falling so

t=(25+√497)/32

t≈1.48 seconds (to nearest hundredth of a second)