Question 678584
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