an astronaut standing on the surface of the moon throws a rock with an initial velocity of 27 feet per second .the astronaut's hand is 6 feet above the surface of the moon , the height of the rock is given by h= -2.7t^2+27t+6.
how many seconds is the rock in the air??
and by the way this problem must solve using quadratic equations and it is :
x= -b+-sequre rode of D /2a
thanx alot
2006-12-19 12:54:46 · 6 answers · asked by pearl 2 in Science & Mathematics ➔ Mathematics
You have basically described all the steps, but you just need to put it together.
You have the formula for the parabola:
h(t) = -2.7t² + 27t + 6
And you want to know when the rock hits the ground. In other words, what is the value of t when the height h(t) is zero?
Start with your parabola formula and set it to zero.
h(t) = -2.7t² + 27t + 6 = 0
Here you have at² + bt + c = 0
with,
a = -2.7
b = 27
c = 6
So by the quadratic formula:
t = [ -b ± sqrt( b² - 4ac ) ] / 2a
t = [ -27 ± sqrt( (-27)² - 4(-2.7)(6) ) ] / 2(-2.7)
t = [ -27 ± sqrt( 729 - (-64.8) ) ] / -5.4
t = [ -27 ± sqrt( 793.8 ) ] / -5.4
t = [ -27 ± 28.1744565 ] / -5.4
t = -0.2175
or
t = 10.2175
Obviously we don't care about the negative time, so the answer is that the rock was in the air for approximately 10.22 seconds.
2006-12-19 12:59:58 · answer #1 · answered by Puzzling 7 · 1⤊ 1⤋
t=-b/(2a) is the vertex which is the time at which the rock will be at its highest. When it's at the vertex, it's halfway done so mutiplying this number by 2 will give you the total time the rock is in the air (half up, half down). t=-27/(2*-2.7)=5 sec. So the answer is 10 seconds.
By the way, you should set h=6, not 0 if you plan on using the quadratic formula since that is the initial height.
So 6=-2.7t^2+27t+6 implies that 0=-2.7t^2+27t and use the quadratic formula from here by letting a=-2.7, b=27, c=0.
2006-12-19 21:00:47 · answer #2 · answered by Professor Maddie 4 · 0⤊ 0⤋
The quadratic formula finds your x intercepts-- that is when the rock is on the ground and when it hits the ground. Since the astronaunt and the rock is above the ground to begin with, the rock is never on the ground for a positive time; therefore only one answer will make sense.
( -27 +/ sqrt[ (27^2) - 4[-2.7 * 6] ] ) / (2 * -2.7)
your two answers are: -.2174919475 or 10.21749195
Since the negative answer doesn't make sense, the rock is in the air for approximately 10.217 seconds
2006-12-19 21:16:55 · answer #3 · answered by j 4 · 0⤊ 0⤋
Put h = 0 and solve as it is the final position , only one of the two solutions will be correct, select which ever is logical
ie
Solve
0= -2.7t^2+27t+6.
2006-12-19 21:03:48 · answer #4 · answered by Alavalathi 3 · 0⤊ 1⤋
Substitute h = 0 into the formula. You will get two answers using the quadratic formula. However one will be inadmissible as it will be negative and time must be greater than 0.
2006-12-19 20:59:31 · answer #5 · answered by keely_66 3 · 0⤊ 1⤋
Looks like you have plenty of help already, you might want to save this for future reference... http://www.algebrahelp.com/calculators/equation/calc.do?equation=%28m0.5+%2B+33%29+%2B+33%3D+108&solvf=AUTO
2006-12-19 21:05:17 · answer #6 · answered by Papa 7 · 0⤊ 0⤋