SOLUTION: A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s.
When will the ball reach a height of 80 m?
Question 117899: A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s.
When will the ball reach a height of 80 m?
You can put this solution on YOUR website! A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s.
When will the ball reach a height of 80 m?
:
I assume they mean 80 m above the ground so it will at 80 m it, be 20 m below the top of the building.
:
The equation:
h = -4.9t^2 + 20t + 100
Where
h is the height of the ball after t seconds
:
-4.9t^2 is the force of gravity, negative because it pulls downward
+20x is the velocity of the ball upward
100 is the initial height when the ball is thrown
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To solve:
-4.9t^2 + 20t + 100 = 80 m
:
-4.9t^2 + 20t + 100 - 80 = 0
:
-4.9t^2 + 20t + 20 = 0; a quadratic equation that we can solve
:
Use the quadratic formula to solve this a=-4.9; b = 20; c = 20
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I got the positive solution of 4.9 seconds (rounded off)
:
:
Check solution in -4.9t^2 + 20t + 100 = h
-4.9(4.9^2) + 20(4.9) + 100 =
-4.9(24) + 98 + 100 =
-117.6 + 98 + 100 = 80.4 ~ 80
