Asked by Court
the puma, which can jump to a height of 12.9 ft when leaving the ground at an angle of 44°. With what speed, in SI units, must the animal leave the ground to reach that height?
Answers
Answered by
bobpursley
The vertical component of velocity must be
Vf^2=2 g 12.9 where g=32ft/sec^2
Vf= sqrt(2*32*12.9)= 28ft/sec
so the velocity of the animal then is Vf/sin44 or about 28*1.4
you can do it more accurately.
Vf^2=2 g 12.9 where g=32ft/sec^2
Vf= sqrt(2*32*12.9)= 28ft/sec
so the velocity of the animal then is Vf/sin44 or about 28*1.4
you can do it more accurately.
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