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You & a friend are hiking in the mountains. You want to climb a ledge that is 20 feet above you. The height of the grappling hook can be thrown is given by the function h(t)=-16t^2+32t. What is the maximum height the grappling hook can reach?

Respuesta :

Answer:

The maximum height the grappling hook can reach with given height function is 16 feet .

Step-by-step explanation:

Given  as :

The height of the grappling hook thrown as function of h(t) = - 16 t² + 32 t

For maximum height ,

[tex]\frac{\partial h(t)}{\partial t}[/tex] = 0

Or, [tex]\frac{\partial ( -16t^{2}+32t)}{\partial t}[/tex] = 0

Or, - 32 t + 32 = 0

Or, 32 t = 32

∴         t = [tex]\frac{32}{32}[/tex] = 1

So, The maximum height is at t = 1 , h(t) =   - 16 t² + 32 t

I.e  h = - 16 ( 1 )² + 32 (1)

Or, h = - 16 + 32

∴    h = 16 feet

Hence The maximum height the grappling hook can reach with given height function is 16 feet .   Answer  

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