SOLUTION: A stone is thrown upwards with an initial speed of 56 m/s. its height is h metres above the ground aft t seconds is given approximately by the formula h= 56t-5t^2. After how many s

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A stone is thrown upwards with an initial speed of 56 m/s. its height is h metres above the ground aft t seconds is given approximately by the formula h= 56t-5t^2. After how many s      Log On


   

Question 776285: A stone is thrown upwards with an initial speed of 56 m/s. its height is h metres above the ground aft t seconds is given approximately by the formula h= 56t-5t^2. After how many seconds will the stone be 50m above the ground? Explain.
Thats the question, i would truly appreciate it if someone would help me clarify this question... :)
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+h+=+56t+-+5t%5E2+
The first term, +56t+ has nothing to do with gravity,
just the upward velocity of the stone
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The 2nd term has only to do with the downward
force of gravity
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The stone will be at a height of +50+ m on its
way up and also again on its way down
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Solve for +h+=+50+
+50+=+56t+-+5t%5E2+
+-5t%5E2+%2B+56t+=+50+
Divide both sides by +-5+
+t%5E2+%2B+%28-56%2F5%29%2At+=+-10+
Complete the square:
+t%5E2+-+%2856%2F5%29%2At+%2B+%28+28%2F5+%29%5E2+=+-10+%2B+%28+28%2F5+%29%5E2+
+t%5E2+-+%2856%2F5%29%2At+%2B+%28+28%2F5+%29%5E2+=+-250%2F25+%2B+784%2F25+
+t%5E2+-+%2856%2F5%29%2At+%2B+%28+28%2F5+%29%5E2+=+534%2F25+
+%28+t+-+28%2F5+%29%5E2+=+534%2F25+
Take the square root of both sides
+t+-+28%2F5+=+-23.108%2F5+
+t+=+%28+28+-+23.108+%29+%2F+5+
+t+=+.978+
And also, taking the positive square root
of the right side:
+t+-+28%2F5+=+23.108%2F5+
+t+=+51.108%2F5+
+t+=+10.222+
The rock is at a height of 50 m at .978 sec
and also at 10.222 sec
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Here's the plot:
+graph%28+400%2C+400%2C+-2%2C+12%2C+-20%2C+170%2C+-5x%5E2+%2B+56x+%29+
The points ( .978, 50 ) and ( 10.222, 50 )
look very possible. Hope this helps.