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.

Question

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.

Answer 1

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