height of object moving

kggirl

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Joined
Oct 5, 2005
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43
The height of an object moving vertically is given by s = -4.9t^2 + 100t. Find its maximum height when it occurs.
 
ds/dt = -9.8t+100 = 0
solve for t and plug it into the original equation for max height.
 
Gene said:
ds/dt = -9.8t+100t = 0
solve for t and plug it into the original equation for max height.

Is this right:

2(-4.9) + 100(1)=
-9.8t + 100=
0=-9.8t + 100

-100/-9.8 = -9.8t/-9.8
10.20 = t

-9.8(10.20)+100(-9.8) = .408 ft
 
Gene wrote:
ds/dt = -9.8t+100t = 0
Not true!
Gene wrote:
ds/dt = -9.8t+100 = 0

You are fine up to the last line where t = 10.2 and the original equation
s = -4.9t^2 + 100t becomes
-4.9(10.2)^2+100(10.2)
 
Gene said:
Gene wrote:
ds/dt = -9.8t+100t = 0
Not true!
Gene wrote:
ds/dt = -9.8t+100 = 0

You are fine up to the last line where t = 10.2 and the original equation
s = -4.9t^2 + 100t becomes
-4.9(10.2)^2+100(10.2)


OK, it should be -9.8(104.04)+90.2=
-1019.592+90.2 =
-929.392
 
just wanted to add that your answer should be in units of m since 9.8m/s^2 is the acceleration of gravity in metric units, not feet. noticed you used the ft units somewhere up there.... its makes a difference being that if it was in units of feet your original function would have looked like f(x)=-16t^2+100t ... just a little physics on the side.
 
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