SOLUTION: You and a friend are hiking in the mountains. You want to climb to a ledge that is 20 ft. above you. The height of the grappling hook you throw is given by the function h(t) = -

Question 678584: You and a friend are hiking in the mountains. You want to climb to a ledge that is 20 ft. above you.
The height of the grappling hook you throw is given by the function
h(t) = -16t^2− 32t + 5 . What is the
maximum height of the grappling hook? Can you throw it high enough to reach the ledge?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The height of the grappling hook you throw is given by the function
h(t) = -16t^2 - 32t + 5 . What is the maximum height of the grappling hook?
Can you throw it high enough to reach the ledge?
:
No, the -32t means it is thrown downward at 32 ft/sec. Upward would be + 32t
:
Assuming the Equation should be -16t^2 + 32t + 5
h = -16t^2 + 32t + 5
find the vertex by finding the axis of symmetry first
t = -32/(2*-16)
t = -32/-32
t = +1 second to reach max height
:
Find the height
h = -16(1^2) + 32(1) + 5
h = -16 + 32 + 5
h = 21 ft is the max height so seems like you would be able to
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