A: To find the squirrel's max height, we need to determine the vertex of the parabolic equation h= -2t^2 + 9t + 11. The formula for the x-coordinate of the vertex is given by t = -b/2a. In this case, a = -2 and b = 9, so t = -9 / 2(-2) = -9 / -4 = 2.25 seconds.
To find the max height, substitute t = 2.25 into the equation h= -2t^2+9t+11:
h = -2(2.25)^2 + 9(2.25) + 11
h = -2(5.0625) + 20.25 + 11
h = -10.125 + 20.25 + 11
h = 21.125 feet
Therefore, the squirrel's max height was 21.125 feet.
B: The squirrel reaches this max height at t = 2.25 seconds.
C: To find when the squirrel reaches the ground, we need to set h = 0 and solve for t in the equation h= -2t^2+9t+11:
0 = -2t^2 + 9t + 11
Rearranging:
2t^2 - 9t - 11 = 0
Using the quadratic formula t= (-(-9) +/- sqrt((-9)^2 - 4(2)(-11))) / 2(2):
t = (9 +/- sqrt(81 + 88)) / 4
t = (9 +/- sqrt(169)) / 4
t = (9 +/- 13) / 4
There are two possible solutions:
1) t = (9 + 13) / 4 = 5.5 seconds
2) t = (9 - 13) / 4 = -1 second
Since time cannot be negative, the squirrel reaches the ground at 5.5 seconds after jumping from the tree.
A flying squirrel jumped from a tree 11 feet in the air at an initial velocity of 9 feet per second. The equation h= -2t^2+9t+11 models his jump where h is height in feet and t is time in seconds.
A: What was the squirrel's max height?
B: How many seconds after he jumped was the squirrel ad this highest point?
C: When did the squirrel reach the ground?
1 answer