Question

A projectile is fired into the air with an initial vertical velocity of 160 ft/sec from ground level. How many seconds later does the projectile reach the maximum height? (numerical answer only)
h(t)=−16t2+160t

Answer 1

The maximum height of the projectile is the maximum point that can be gotten from the projectile equation

The projectile reaches the maximum height after 5 seconds

The function is given as:

$\mathbf{h(t) = -16t^2 + 160t}$

Differentiate the function with respect to t

$\mathbf{h'(t) = -32t + 160}$

Set to 0

$\mathbf{h'(t) = -32t + 160 = 0}$

So, we have:

$\mathbf{-32t + 160 = 0}$

Collect like terms

$\mathbf{-32t =- 160 + 0}$

$\mathbf{-32t =- 160}$

Solve for t

$\mathbf{t = \frac{- 160}{-32}}$

$\mathbf{t = 5}$

Hence, the projectile reaches the maximum after 5 seconds

Read more about maximum values at:

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