SOLUTION: Could someone help me please? The problem is: A ball is thrown upward from the roof of a building 100m tall with an initial velocity of 20m/s. When will the ball reach a height of

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Question 87894: Could someone help me please? The problem is:
A ball is thrown upward from the roof of a building 100m tall with an initial velocity of 20m/s. When will the ball reach a height of 80m?
Found 2 solutions by Flake, Earlsdon:
Answer by Flake(45) About Me  (Show Source):
You can put this solution on YOUR website!
This borders on Physics questions. There are many ways to solve Physics problems. So I'll make some assumptions that you know about the steps I'm listing out.
****Step#1: Start with general Free-fall equation.
d+=+vt+-+%281%2F2%29gt%5E2+ ; where d = distance; v = velocity, t = time
****Step#2: draw a picture
You're on a roof tome 100m from the ground.
you throw the ball up with initial speed of 20m/s
Now imagine the path.....the ball moves up. Reaches max height and start to come back down.
You should have drawn an upside-down parabola.
****Step#3: Solving for the time the ball travelles to the top and returns to the same height that it left your hand.
----Take the derivative of the equation to find time to reach top, because derivative is to find the Vertex of a parabola
d+=+vt+-+%28g%2F2%29t%5E2+
Derivative becames 0+=+v+-+2%28g%2F2%29t+
becames 0+=+v+-+gt+
becames gt+=+v+
becames t+=+v%2Fg+=+2.04+ where v=20 m/s and g = 9.8 m/(s^2)
(Since the ball travels in parabola, the time it takes to reach the top is the same as the time it takes to return your hand.)
----Total time the ball took to leave your hand at 100m from and return to same 100m height should be 4.08 sec.
-----Step#4: Calculate the time took to travel from 100m to 80m. Note, this is when you don't catch it and let it continue to fall down to 80m.
-----the equation is SIMILAR.
d+=+vt+%2B+%28g%2F2%29t%5E2+ , Notice the "+" instead of the "-" in front of the g/2. This is because gravity is now working to speed up the ball as it falls. The first case, gavity is slowing the ball down as it was travelling up, so the "-" in the first case.
-----Now the ball travells down from 100m to 80m, so it moved 20m.
d+=+vt+%2B+%28g%2F2%29t%5E2+; where d = 20, (note that V=20m/s, reason is at 100m the ball is should travelling at the same speed down as when it was thrown from your hand up.)
becomes 20+=+20t+%2B+%289.8%2F2%29t%5E2+
becomes %289.8%2F2%29t%5E2+%2B+20t+-+20+=+0
Solve with x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
--Give t = 0.83 sec
--------Step#5: Total time to get to 80m from the moment you throw the ball up at 100m.
4.08+0.83 = 4.91 sec
---------hope that helped.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, you can start with the function that gives the height, as a function of time, h(t) of an object propelled upward with an initial upward velocity of v%5B0%5D%5D%5Dm/s from an initial height of h%5B0%5Dm.
h%28t%29+=+-4.9t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D
In this problem:
h%5B0%5D+=+100m.
v%5B0%5D+=+20m/s.
You would like to know at what time, t, will h = 80 meters.
Making the appropriate substitutions into the formula:
80+=+-4.9t%5E2%2B20t%2B100 Subtract 80 from both sides.
-4.9t%5E2%2B20t%2B20+=+0 Solve this quadratic equation using the quadratic formula: t+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a In this equation, a = -4.9, b = 20, and c = 20
t+=+%28-20%2B-sqrt%2820%5E2-4%28-4.9%29%2820%29%29%29%2F2%28-4.9%29 Simplifying this, you'll get:
t+=+%28-20%2B-sqrt%28400-%28-392%29%29%29%2F-9.8
t+=+%28-20%2B-sqrt%28792%29%29%2F-9.8 Evaluating this gets you:
t+=+4.912499 or t+=+-0.830866 Discard the negative solution.
t+=+4.9seconds.