The best leaper in the animal kingdom is the puma, which can jump to a height of 12.2 ft when leaving the grou

Question

The best leaper in the animal kingdom is the puma, which can jump to a height of 12.2 ft when leaving the ground at an angle of 43°. With what speed, in SI units, must the animal leave the ground to reach that height?
m/s

Answer 1

conservatin of energy is the most direct way to solve this

set a variable vy=vi*sin(43)

g*h=.5*vy^2
vy=sqrt(2*32*12.2)
=28 ft/second

vi=28/sin(43)
=41 ft/sec

just for fun, this can be derived using the motion equations
let vyi=vi*sin(43)
vy(t)=velocity in the y direction
t is time with t=0 the jump
y(t) is position in the y direction

y(t)=vyi*t-.5*g*t^2
vy(t)=vyi-g*t

when vy(t)=0, the puma reaches maximum height of 12.2 ft

0=vyi-g*t
t=vyi/g

12.2=vyi*t-.5*g*t^2
12.2=vyi^2/g-.5*vyi^2/g
vyi=sqrt(12.2*g*2)

same result


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