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
8.4 = u t = .920 S t
S t = 9.13
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