Question 1212: An object is projected upward with an initial velocity of 98ft/second. The rise and fall of the object is given by the equation s=98t-4.9t2, where s is the height above the ground and t is the time from the object's launch. Find the maximum height and the time this occurs?
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! s(t)=98t-4.9t^2,
s'(t) = 98 - 9.8 t,
Solve s'(t) = 0, i.e. 98 - 9.8 t = 0, or 9.8 t = 98,
we have t = 10 sec.
Since s"(t) = -9.8 < 0 for all t, by the 2nd derivative test,
s has maximum value when t = 10 sec.
We see that the maximum height is s(10) = 980 - 4.9 *100 = 490 meters.
If you don't know derivative.
Use s(t)=98t-4.9t^2 = -4.9(t^2 - 20t + 100) + 490
= -4.9(t - 10)^2 + 490 (<= 490)
We can see that s(t) has max value 490 when t = 10.
Or use the formula v(velocity) = 98 - 9.8 t.
If v =0, the height is max.
Solve v = 98 - 9.8 t = 0, we get t = 10 sec
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