SOLUTION: Hello!
If a ball is thrown directly upward with a velocity of 31 ft/s, its height (in feet) after t seconds is given by y = 31t - 16t^2. What is the maximum height attained by t
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Question 199422: Hello!
If a ball is thrown directly upward with a velocity of 31 ft/s, its height (in feet) after t seconds is given by y = 31t - 16t^2. What is the maximum height attained by the ball?
thanks!
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Using the model you provided (which, by the way, is a very poor model for a ball being thrown straight upward at such a low initial velocity -- more to follow), you first need to set your function into standard form, that is 
.
 = -16t^2 + 31t)
Notice that this is the equation of a parabola. The lead coefficient is less than zero, so it opens downward. Therefore, the vertex is a maximum value for the function. The independent variable coordinate of the vertex of any parabola of the form is given by 
, hence:
}= \frac{31}{32})
,
and the value of the function at that time value is:
 = -16\left(\frac{31}{32}\right)^2 + 31\left(\frac{31}{32}\right))
You get to do your own arithmetic.
The reason that your model for this particular situation is such a poor one is that it assumes that the initial height is zero. That means that you are either modeling this based on zero height being the height above the ground at which the thrower's hand was when the ball was released (a very odd choice for a height baseline indeed) or somehow this person was able to release the ball straight upward at 31 feet per second while their hand was touching the ground. If you look at the numbers here, the time at max height is just a little less than 1 second, meaning that the height calculated by the model is going to be very nearly 15 feet. But a person in the height range of 5 to 6 feet tall is going to release a ball thrown straight up at about 6 to 7 feet above the ground. That changes the actual height reached by this particular thrown ball by something on the order of 40%. The correct height model, neglecting atmospheric friction effects, is:
 = -\frac{1}{2}gt^2 + v_ot + h_o)
Where 
or 
depending on the system of units and assuming you are on planet Earth, 
is the initial velocity, and 
is the initial height.
John
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