Question

a waterfall has a height of 1300 feet. a pebble is thrown upward from the top of the falls with an initial velocity of 20 feet per second. the height, h, of the pebble after t seconds is given by the equation ( h = -16t^2 + 20t + 1300 ). how long after the pebble is thrown will it hit the ground?

the pebble will hit the ground about \boxed{} seconds after it is thrown.

(simplify your answer. round to one decimal place as needed)

Explanation:

Step1: Set h = 0 (ground level)

We need to solve the quadratic equation \( -16t^2 + 20t + 1300 = 0 \). First, multiply both sides by -1 to make the coefficient of \( t^2 \) positive: \( 16t^2 - 20t - 1300 = 0 \).

Step2: Use quadratic formula

The quadratic formula is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 16 \), \( b = -20 \), and \( c = -1300 \). Plugging in these values:
First, calculate the discriminant \( D = b^2 - 4ac = (-20)^2 - 4(16)(-1300) = 400 + 83200 = 83600 \).
Then, \( t = \frac{20 \pm \sqrt{83600}}{32} \). We take the positive root (since time can't be negative), so \( t = \frac{20 + \sqrt{83600}}{32} \).
Calculate \( \sqrt{83600} \approx 289.1367 \). Then \( t = \frac{20 + 289.1367}{32} \approx \frac{309.1367}{32} \approx 9.6605 \). Rounding to one decimal place, \( t \approx 9.7 \).

Answer:

9.7