Question

An object is thrown. t seconds after it is thrown, its height is given by h(t) = -16t^2 + 160t + 400 feet The object reaches its highest point after seconds. The object's maximum height is feet. 

          An object is thrown. t seconds after it is thrown, its height is given by

h(t) = -16t^2 + 160t + 400 feet

The object reaches its highest point after seconds.

The object's maximum height is feet.
An object is thrown. t seconds after it is thrown, its height is given by h(t) = -16t^2 + 160t + 400 feet The object reaches its highest point after seconds. The object's maximum height is feet.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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An object is thrown. t seconds after it is thrown, its height is given by h(t) = -16t^2 + 160t + 400 feet. The object reaches its highest point after seconds. The object's maximum height is feet.
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Transcript

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00:01 For this problem we are giving the following we are giving that the height of some object of an object is described or given by h of t where h of t is equal to negative 16 t squared plus 160t plus 400 and the units for the height are measured in feet uh next we want to find first for part a when the object reaches its highest point so the object reaches its highest point highest point when t is equal to negative b over 2a so in our case it is negative 160 the constant value b is the coefficient of t divided by 2a where a is negative 16th this is just going to equal to 5 and the units for this one…
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