Question

A waterfall has a height of 1400 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 12 feet per second. The​ height, h, of the pebble after t seconds is given by the equation h equals negative 16 t squared plus 12 t plus 1400

h=−16t2+12t+1400. How long after the pebble is thrown will it hit the​ ground?

Answer 1

To determine when the pebble will hit the ground, we set the height equation equal to 0 and solve for t.

Step-by-step explanation:

To determine when the pebble will hit the ground, we set the height equation equal to 0. The equation is:

h = -16t^2 + 12t + 1400

Setting h equal to 0, we can solve for t:

0 = -16t^2 + 12t + 1400

We can solve this quadratic equation by factoring or using the quadratic formula. By factoring, we can rewrite the equation as:

0 = -4(4t^2 - 3t - 350)

From here, we can either try factorizing further or use the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

Using this formula, we can find the values of t when the pebble hits the ground.

Answer 2

I think the answer is 10