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
The formula is
s = sO + vOt + atē/2
where
s = h = the height of the ball off the ground in meters
sO = hO = the initial height = 100 m
vO = the initial velocity = 20 m/s
t = the number of seconds
a = g = -9.8 m/sē (the acceleration of gravity)
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Substituting:
h = (100) + (20)t + (-9.8)tē/2
h = 100 + 20t - 4.9tē
That's the formula for the height h in meters
at any time t seconds.
Plug in h = 80m and solve for t
80 = 100 + 20t - 4.9tē
4.9tē - 20t - 20 = 0
If you like, clear of decimals by multiplying
thru by 10
49tē - 200t - 200 = 0
Use the quadratic formula and you'll get
t = -.83 and 4.92, approximately
Discard the negative answer. The legitimate answer is
4.92 seconds.
Analysis:
at 0 seconds, the height is 100 m
at 1 second, the height is 115.1 m
at 2 seconds the height is 120.4 m
then it starts falling
at 3 seconds, its height is 115.9 m
at 4 seconds its height is 101.6 m
at 4.92 seconds its height is 80m
at 5 seconds its height is 77.5 m
at 6 seconds its height is 43.6 m
at 7 seconds its height is 0, i.e., it hits the ground.
Edwin
