A waterfall has a height of 1100 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 16 feet per second. The height of the pebble h in feet after t seconds is given by the equation h equals negative 16 t squared plus 16 t plus 1100. How long after the pebble is thrown will it hit the ground?
To solve this problem, we need to find the time when the pebble hits the ground, which is when the height of the pebble (h) becomes 0.
Given information: - The height of the waterfall is 1100 feet. - The initial velocity of the pebble is 16 feet per second. - The height of the pebble (h) in feet after t seconds is given by the equation: h = -16t^2 + 16t + 1100
Step 1: Set the height of the pebble (h) equal to 0 to find the time when it hits the ground. 0 = -16t^2 + 16t + 1100
Step 2: Solve the quadratic equation for t. -16t^2 + 16t + 1100 = 0 t^2 - t - 68.75 = 0
Using the quadratic formula: t = (-(-1) ± √((-1)^2 - 4 × (-16) × 1100)) / (2 × (-16)) t = (1 ± √(1 + 70400)) / (-32) t = (1 ± √70401) / (-32)
There are two solutions for t, but we are interested in the positive value, which represents the time when the pebble hits the ground.
t = (1 + √70401) / (-32) t = (1 + 265.33) / (-32) t = 266.33 / (-32) t = -8.32 seconds
Therefore, the pebble will hit the ground 8.32 seconds after it is thrown.
