Asked by anonymous
A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 11 from the takeoff point.
If the kangaroo leaves the ground at a 18 angle, what is its takeoff speed ?
What is its horizontal speed?
If the kangaroo leaves the ground at a 18 angle, what is its takeoff speed ?
What is its horizontal speed?
Answers
Answered by
Melissa
Takeoff Speed= Range=(initial velocity^2 * Sin (2 theta))/g
11m = (V0^2*sin(36))/(9.8)
V0= sqrt((11*9.8)/(sin36))
V0=13.54
11m = (V0^2*sin(36))/(9.8)
V0= sqrt((11*9.8)/(sin36))
V0=13.54
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