Posted by **Yuki** on Thursday, September 27, 2012 at 6:37pm.

A gray kangaroo can bound across level ground with each jump carrying it 8.4m from the takeoff point. Typically the kangaroo leaves the ground at a 23 angle.

1. Provide knowns and unknown values.

2. What is the initial velocity or takeoff speed?

3. What is its maximum height above the ground?

Please provide formula and numerical answer.

- Physics -
**Damon**, Thursday, September 27, 2012 at 7:06pm
say speed is S

then horizontal velocity u = S cos 23

Vi, initial vertical velocity = S sin 23

t is total time in air and t/2 is time to max height

d = distance = u t

so

8.4 = u t = .920 S t

S t = 9.13

at top

time = t/2 = 9.13/2S = 4.57/S

v = 0 = Vi - 9.8 (4.57/S)

0 = .391 S - 44.8/S

.391 S^2 = 44.8

S = 10.7 m/s takeoff speed

h = Vi (t/2) - 4.9 (t/2)^2

t/2 = 4.57/S = .427

Vi = .391 (10.7) = 4.18

h = 4.18(.427) - 4.9 (.427)^2

h = .891

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