Posted by Paul on Monday, August 6, 2012 at 9:14pm.
The acceleration of an object g = -9.8 m/s^2. From the top of a building of 245 m., an object is thrown upwards at an initial velocity of 12m/s
(a) What is the function for the velocity of the object?
I put vf^2 = Vi^2 + 2a(yf-yi)
vf^2 = 216.09 + 19.6 (yf - 245)
(b) Determine the function for the position of the object relative to the ground.
I don't know what to put for this one
(c) How long will it take for the object to reach maximum height?
I know that for maximum height velocity = 0 with the formula: Y = Yo + Vot+ 1/2at^2, but the problem is that when I try to isolate t, I still have the t from Vot which I don't know where to put, so I'm left with t = squ(2(Yo + Vot - Y))/a
(d) What is the velocity of the object once it reaches the ground?
For this one I'll use the same formula as for (a) but will my initial height be the maximum height reached from (c)?
- Math - Reiny, Monday, August 6, 2012 at 10:25pm
my equation would be
s(t) = -4.9t^2 + 12t + 245
v(t) = -9.8t + 12
at the maximum height, v = 0
-9.8t + 12 = 0
t = 12/9.8 = 1.22 seconds
when it hits the ground, s(t) = 0
-4.9t^2 + 12t + 245 = 0
4.9t^2 - 12t - 245 = 0
using the formula ...
t = 8.4 or a negative
so at 8.4 seconds, v(8.4) = -9.8(8.4) + 12 = -70.32 m/sec
- Math - Robert, Monday, August 6, 2012 at 10:39pm
so for (a) you used Y = Yo + Vot+ 1/2at^2 so when you derive it, you get time for the other ones, got it. THanks a lot Reiny
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