SOLUTION: Using the formula h=-1/2gt^2+v(0)t+h(0) (the zero in the parens is lower/small beneath the letter) and g= gravitational constant and v = upward velocity from a starting height of h

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Using the formula h=-1/2gt^2+v(0)t+h(0) (the zero in the parens is lower/small beneath the letter) and g= gravitational constant and v = upward velocity from a starting height of h      Log On


   

Question 1005463: Using the formula h=-1/2gt^2+v(0)t+h(0) (the zero in the parens is lower/small beneath the letter) and g= gravitational constant and v = upward velocity from a starting height of h.
1a. What is the height after three seconds of a penny dropped from the roof of the a tall building about 1250 feet tall?
1b. After how many seconds does the penny hit the ground?
2. What is the height after three seconds of a ball thrown from the roof of a building about 495 feet tall and is tossed upwards with a velocity of 10m/s?
2b. After how many seconds does it hit the ground?
Thanks in advance. I think I have the answers but want to check the work done.
Found 2 solutions by KMST, rothauserc:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
h=-%281%2F2%29gt%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D is a general formula.

For the first problem we will set:
t= time from the moment the penny is dropped (we will measure it in seconds, because it will be a short time),
h%5B0%5D=1250= height (in feet) at the moment, t=0 , when the penny is dropped,
v%5B0%5D=0= the initial velocity (in feet/second) of the penny at t=0 (since it was "dropped", it was not thrown up or down with any initial velocity).
We will use the acceleration of gravity in ft%2Fsecond%5E2 , to agree with the other units:
g=32 = acceleration of gravity in ft%2Fsecond%5E2 .
Substituting, we get
h=-%281%2F2%2932t%5E2%2B0%2At%2B1250--->h=-16t%5E2%2B1250

1a. For t=3 , h=-16%2A3%5E2%2B1250=-16%2A9%2B1250=-144%2B1250=1106 ,
so the height of the penny after three seconds is highlight%281106feet%29 .
1b. When the penny hits the ground, h=0 . Then,
0=-16t%5E2%2B1250--->16t%5E2=1250--->t%5E2=1250%2F16--->t%5E2=78.125--->t=sqrt%2878.125%29--->t=about+8.84 ,
so the penny hits the ground after about highlight%288.85seconds%29 .


For the second problem we will set:
t= time from the moment the ball is tossed (we will measure it in seconds, because it will be a short time),
v%5B0%5D=10= the initial velocity (in meters/second) of the ball at t=0 (since it was "dropped", it was not thrown up or down with any initial velocity).

red%28IF%29 (as I originally misread) h%5B0%5D=495= height (in meters) at the moment, t=0 , when the ball is tossed,
We will use the acceleration of gravity in meters%2Fsecond%5E2 , to agree with the other units:
g=9.8 = acceleration of gravity in meters%2Fsecond%5E2 .
Substituting, we get
h=-%281%2F2%299.8t%5E2%2B10t%2B495--->h=-4.9t%5E2%2B10t%2B495
2. For t=3 ,
h=-4.9%2A3%5E2%2B10%2A3%2B495-->h=-4.9%2A9%2B30%2B495-->h=-44.1%2B30%2B495-->h=480.9 ,
so the height of the ball after three seconds is highlight%28480.9meters%29 .
2b. When the ball hits the ground, h=0 . Then,
0=-4.9t%5E2%2B10t%2B495 .
Solving the quadratic equation we get two solutions, but the negative solution does not make sense, and needs to be discarded.
So, t=%28-10+%2B+sqrt%2810%5E2-4%2A%28-4.9%29%2A495+%29%29%2F%282%2A4.9%29
t=%28-10+%2B+sqrt%28100%2B9702%29%29%2F9.8
t=%28-10+%2B+sqrt%289802%29%29%2F9.8
t=about%28-10+%2B+99%29%2F9.8=89%2F9.8=about9.08 ,
so the ball hits the ground after about highlight%289.08seconds%29 .

red%28IF%29 it was really meant as h%5B0%5D=495= height (in feet),
we have a units mismatch.
We have to get everything in feet, or everything in meters.
Since I am a supporter of the SI system of units, I will convert 495+feet to meters.
495+feet=%28495+feet%29%280.3048meters%2F%221+foot%22%29=150.9m(rounded), so
h%5B0%5D=150.9= height (in meters) at the moment, t=0 , when the ball is tossed,
We will use the acceleration of gravity in meters%2Fsecond%5E2 , to agree with the other units:
g=9.8 = acceleration of gravity in meters%2Fsecond%5E2 .
Substituting, we get
h=-%281%2F2%299.8t%5E2%2B10t%2B150.9--->h=-4.9t%5E2%2B10t%2B150.9
2. For t=3 ,
h=-4.9%2A3%5E2%2B10%2A3%2B150.9-->h=-4.9%2A9%2B30%2B150.9-->h=-44.1%2B30%2B495-->h=136.8 ,
so the height of the ball after three seconds is highlight%28136.8meters%29 .
2b. When the ball hits the ground, h=0 . Then,
0=-4.9t%5E2%2B10t%2B150.9 .
Solving the quadratic equation we get two solutions, but the negative solution does not make sense, and needs to be discarded.
So, t=%28-10+%2B+sqrt%2810%5E2-4%2A%28-4.9%29%2A150.9+%29%29%2F%282%2A4.9%29
t=%28-10+%2B+sqrt%28100%2B2957.6%29%29%2F9.8
t=%28-10+%2B+sqrt%283057.6%29%29%2F9.8
t=about%28-10+%2B+55.30%29%2F9.8=45.30%2F9.8=about4.62 ,
so the ball hits the ground after about highlight%284.62seconds%29 .

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
note when dropping an object from height, the initial velocity vo is 0. G is 16 for this problem, since given values are in feet.
*********************************************************************************
1a) h(3) = -8*3^2 + 0*3 + 1250 = 1178
after 3 seconds, the penny is 1178 feet above the ground
1b) 0 = -8t^2 +1250
t^2 = 156.25 and t = 12.5
after 12.5 seconds, the penny hits the ground
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the height of the building is given in feet, so convert vo = 10 m to feet/s, vo = 32.8 feet
2a) h(3) = -8*3^2 + 32.8*3 + 495 = 521.4
after three seconds, the ball is 521.4 feet above the ground.
2b) 0 = -8t^2 + 32.8t + 495
t^2 -4.1t - 61.875 = 0
t^2 -4.1t = 61.875
t^2 -4.1t + 4.2025 = 4.2025 + 61.875
(t - 2.05)^2 = 66.0775
t - 2.05 = 8.128806801 = approx 10.2
in 10.2 seconds the ball hits the ground
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here is the graph of 2b
graph%28300%2C+200%2C+-10%2C+12%2C+-2%2C+800%2C+-8x%5E2+%2B32.8x+%2B495%29