Question

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?

Answer 1

Answer: 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:

$t = \frac{-24 +- \sqrt{24^2 - 4*(-16)*1200} }{2*-16} = \frac{-24 +- 278.2}{-32}$

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