Question

A kangaroo jumps off the ground with an initial velocity of 18 feet per second.

a) Write an equation that gives the height (in feet) of the kangaroo as a function of time (in seconds) since it jumps.

b) After how many seconds does that kangaroo land on the ground?

Answer 1

@freckles can you help me with this?

Answer 2

so we know we have a parabola type graph

Answer 3

we are given the initial height is 0 ft
and the initial velocity is 18 ft/sec

Answer 4

ok

Answer 5

h(t)=at^2+bt+c
We have that c is the initial height
b is the initial velocity
and I forgot with a is
a is usually a set number
a is -16 I think

Answer 6

ok

Answer 7

so would the formula be: y=-16t^2+18t?

Answer 8

that a is the number that has to do with gravity
yeah is it -16 is you are dealing with feet
and -4.9 if you are dealing with meters

Answer 9

and we definitely have feet

Answer 10

yep

Answer 11

ok

Answer 12

the next question
on the ground means the height is 0
correct?

Answer 13

so find t for when y is 0

Answer 14

that is solve 0=-16t^2+18t for t

Answer 15

is it 9/8?

Answer 16

yep you have the the solution t=0 or t=9/8... t=0 sec is right before the kangaroo lifts off and t=1.125 or 9/8 sec is when you have the kangaroo lands back on the ground after his hop

Answer 17

ok awesome... thanks again you r really smart

Answer 18

and of course yes you are looking for the 1.125 solution since it asks after how many sec does he land back on the ground

Answer 19

more examples like this problem can be found here:
http://www.purplemath.com/modules/quadprob.htm

Answer 20

anyways peace and have fun

Answer 21

OK GREAT! THX