Question

A flying squirrel jumped from a tree 11 feet in the air at an initial velocity of 9 feet per sencond. The equation h=-2t^2+9t-11 models his jump where h is height in feet and t is time in senconds.
A: What was the squirrels maximum height? (find the vertex point)
B: How many senconds after he jumped was the squirrel at his highest point?
C: When did the squirrel reach the ground?

Answer 1

Answer:

(a) -0.875m from the tree.

(B) At a time, t=2.25s after it jumped.

(C)At a time, t=4.5s after it jumped .

Step-by-step explanation:

First things first, it is important to note that the squirrel can only have been said to reach its maximum height when its velocity equals zero.

If h=-2t^2+9t-11

Then dh/DT=v=-4t+9

A) Therefore ,at the maximum height, the velocity ,v=0.

That is, 4t=9 , consequently, t=2.25s

Therefore, when t=2.25s

h=-2(2.25)^2 + 9(2.25)-11

The maximum height reached is then h=-0.875m.

This means the vertex point is at (11-0.875)m above the ground.

B) As explained above, the squirrel was at its highest point when, v=0, which occurred 2.25seconds after it jumped.

C) The squirrel reaches the ground when its at an height ,h=-11feets

Therefore, -11=-2t^2+9t-11

Them, 2t^2= 9t

Therefore,2t=9

Therefore, the squirrel reaches the ground at a time t=4.5s.

Answer 2

This must be on the moon as the acceleration due to gravity in this equation must be around 1/8 that on earth. :)  Anyway...

h=-2t^2+9t+11

A)

dh/dt=-4t+9, when velocity, dh/dt=0, it is the maximum height reached

dh/dt=0 only when 4t=9, t=2.25 seconds

h(2.25)=21.125 ft  (21 1/8 ft)

B)

As seen in A), the time of maximum height was 2.25 seconds after the squirrel jumped.

C)

The squirrel reaches the ground when h=0...

0=-2t^2+9t+11

-2t^2-2t+11t+11=0

-2t(t+1)+11(t+1)=0

(-2t+11)(t+1)=0, since t>0 for this problem...

-2t+11=0

-2t=-11

t=5.5 seconds.