SOLUTION: How do I solve this? A projectile is thrown upward so that its distance above the ground after t seconds is h=-13t^2+520t. After how many seconds does it reach its maximum he

Question 668717: How do I solve this?
A projectile is thrown upward so that its distance above the ground after t seconds is
h=-13t^2+520t. After how many seconds does it reach its maximum height?

Found 3 solutions by jenabee, amalm06, ikleyn:
Answer by jenabee(1)   (Show Source): You can put this solution on YOUR website!
h=-13t^2+520t since a= -13, the parabola will open down.
Use -b/2a to find the x-value. This value will represent the number of seconds the projectile will meet its maximum height.
a= -13
b=250
-250/2(-13)= 20
Therefore, in 20 seconds, the projectile will meet its maximum height.
Answer by amalm06(224)   (Show Source): You can put this solution on YOUR website!
The projectile will reach its maximum height when it's velocity is zero.
The velocity is the derivative of the position with respect to time.
In this case, position is denoted by h and given as:
h=-13t^2+520t
Differentiate h with respect to t to obtain the velocity:
v=dh/dt=-26t+520
Now set v equal to zero and solve for t:
-26t+520=0
t=20 sec(Answer)

Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

Double check the number "13" in your post. Usually (as a rule) this number for the Earth conditions is EITHER 16 (when the height is measured in feet) OR 5 (when the height is measured in meters). It is for the first time I see the number "13" in such problems. But if they are talking about a projectile launched at the other planet surface . . . then everything may happen . . .

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