Find All Values Of T For Which The Ball's Height Is 11 Feet

Find all values of t for which the ball's height is 11 feet

The height of the ball \( h(t) \) is given by the formula:

\( h(t) = -16 t^2 + 21 t + 5 \)

We are asked to find all values of \( t \) when the height \( h(t) \) equals 11 feet. This gives us the quadratic equation:

\( -16 t^2 + 21 t + 5 = 11 \)

First, bring all terms to one side to set the equation equal to zero:

\( -16 t^2 + 21 t + 5 - 11 = 0 \)

\( -16 t^2 + 21 t - 6 = 0 \)

Divide the entire equation by -1 to simplify calculations:

\( 16 t^2 - 21 t + 6 = 0 \)

Calculate the discriminant \( \Delta \):

\( \Delta = b^2 - 4ac = (-21)^2 - 4 \times 16 \times 6 = 441 - 384 = 57 \)

Since the discriminant is positive, there are two real solutions. Find them using the quadratic formula:

\( t = \frac{-b \pm \sqrt{\Delta}}{2a} \)

Substituting the known values:

\( t = \frac{21 \pm \sqrt{57}}{2 \times 16} = \frac{21 \pm \sqrt{57}}{32} \)

Calculating the approximate values:

\( t \approx \frac{21 \pm 7.55}{32} \)

First solution:

\( t \approx \frac{21 + 7.55}{32} = \frac{28.55}{32} \approx 0.892 \) seconds

Second solution:

\( t \approx \frac{21 - 7.55}{32} = \frac{13.45}{32} \approx 0.420 \) seconds

Therefore, the ball reaches a height of 11 feet at approximately 0.420 seconds and 0.892 seconds.

References

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