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kvil9368
30.11.2020 •
Mathematics
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?
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Ответ:
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
Ответ:
1 hour and 10 minutes I believe. If you'd like me to explain how I got that I could.