Q4 A basketball is shot from a height of 6 feet with an initial upward velocity of 20 feet per second. The height of the basketball can be represented by the equation h = -16t2 + 20t + 64 where t = time after the shot in seconds and h = the height in feet. When will the basketball hit the ground?

Answer 1

To find when the basketball hits the ground, we need to determine when the height \( h \) is equal to 0. We are given the height equation:

\[ h = -16t^2 + 20t + 64 \]

Setting \( h = 0 \) to find the time \( t \) when the basketball hits the ground, we have:

\[ -16t^2 + 20t + 64 = 0 \]

This is a quadratic equation in the standard form \( at^2 + bt + c = 0 \), where:

• \( a = -16 \)
• \( b = 20 \)
• \( c = 64 \)

We can use the quadratic formula to solve for \( t \):

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Plugging in the values of \( a \), \( b \), and \( c \):

\[ t = \frac{-20 \pm \sqrt{20^2 - 4(-16)(64)}}{2(-16)} \]

Calculating inside the square root:

\[ 20^2 = 400 \] \[ 4(-16)(64) = -4096 \quad \text{(this is actually positive since it's -4 times a negative)} \] Thus: \[ b^2 - 4ac = 400 + 4096 = 4496 \]

Now we take the square root:

\[ \sqrt{4496} \approx 67.03 \]

Now substituting back into the quadratic formula:

\[ t = \frac{-20 \pm 67.03}{-32} \]

Calculating the two possible values for \( t \):

1. Using the positive root:

\[ t = \frac{-20 + 67.03}{-32} \approx \frac{47.03}{-32} \approx -1.47 \quad \text{(not valid, as time cannot be negative)} \]

2. Using the negative root:

\[ t = \frac{-20 - 67.03}{-32} \approx \frac{-87.03}{-32} \approx 2.72 \]

So the basketball will hit the ground approximately \( t \approx 2.72 \) seconds after being shot.