A waterfall has a height of 1200 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 24 feet per second. The height, h, of the pebble after t seconds is given by the equation -16t^2+24t+1200 . How long after the pebble is thrown will it hit the ground?

9.44 seconds

Step-by-step explanation:

A movement equation in the vertical axis is usually written as follows:

p(t) = (-g/2)*t^2 + v0*t + p0

where:

g is the gravitational acceleration, in this case is 32 ft/s^2, then -g/2 = -16ft/s^2

v0 is the initial velocity, in this case, 24 ft/s

p0 is the initial height, in this case, 1200ft

Then, when p(t)  = 0ft will mean that the pebble will hit the ground, then we need to calculate:

p(t) = -16t^2+24t+1200  = 0

and find the value of t, this is a quadratic equation, then we can use the Bhaskara equation to find the two solutions, these are:

Then the two solutions are:

t = (-24  + 278.2)/-32 = -7.94 seconds (we can discard this one, because the negative time is not really defined)

t = (-24 - 278.2)/-32 = 9.44 seconds

Then the pebble needs 9.44 seconds to hit the ground

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.

sorry not knowing about that