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You and a friend are hiking in the mountains. You want to climb to a ledge that is 20ft above you. The height of the grappling hook you throw is given by the function h = - 16t2 – 32t + 5. a) What is the maximum height of the grappling hook? b) Can you throw it high enough to reach the ledge?

 Feb 6, 2015

Best Answer 

 #1
avatar+94558 
+10

I think the function should be h = -16t^2 + 32t + 5

The position of the grappling hook at time 0 = 5 ft.

And the max height it reaches is 21 ft.....but it needs to reach a height of 25ft....so, no, it won't be thrown high enough to reach the ledge....

Here's a graph that substitutes y for h and x for t....

https://www.desmos.com/calculator/pzsjsonnv0

 

 Feb 6, 2015
 #1
avatar+94558 
+10
Best Answer

I think the function should be h = -16t^2 + 32t + 5

The position of the grappling hook at time 0 = 5 ft.

And the max height it reaches is 21 ft.....but it needs to reach a height of 25ft....so, no, it won't be thrown high enough to reach the ledge....

Here's a graph that substitutes y for h and x for t....

https://www.desmos.com/calculator/pzsjsonnv0

 

CPhill Feb 6, 2015
 #2
avatar+3691 
0

good job Chris!!! woop woop!

 Feb 6, 2015
 #3
avatar+94558 
0

HAHAHA, BJ!!!...I just hope it's correct !!! LOL!!!

 

 Feb 6, 2015
 #4
avatar+3691 
0

i believe u are! :P

 Feb 6, 2015

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