The best leaper in the animal kingdom is the puma, which can jump to a height of

Question

The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45 degree. With what speed must the animal leave the ground to reach that height? m/s A jet airliner moving initially at 3.70 times 10 mi/h due east enters a region where the wind is blowing at 1.00 times 10^2 mi/h in a direction 36.0 degree north of east. (Let the x-direction be eastward and the y-direction be northward.) Find the components of the velocity of the Jet airliner relative to the air, Find the components of the velocity of the air relative to Earth, vector v_AB, Write an equation analogous to for the velocities and. (In the given equation, is the velocity of observer B relative to observer E. Object A travels with velocity relative to observer B, and velocity relative to observer E.) What are the speed and direction of the aircraft relative to the ground? magnitude mi/h direction degree north of east

Explanation / Answer

a) vsin45 = sqroot ( 2x 9.8 x 3.7)

v ( 1/ sqroot2) = 8.515

v(initial) = 12.04 m/s apprx

b) vx ( velocity of airliner with respect to air) = 3.70 x 10^ 2 mi/hr- 1x 10^2 cos 36= 2.891 x 10^2 mi/hr apprx

vy ( velocity of airliner with respect to air) = - 1 x 10^2 sin 36 = -0.5877 x 10^2 mi/hr apprx

c) Vx ( velocity of air with respect to earth) = 1 x 10^2 cos 36 = 0.809 x 10^2 mil/hr

VY ( velocity of air with respect to earth)= 0.5877 x 10^2 mi/hr apprx

d) correct option is c

e) VX ( velocity of airliner with respect to ground) = 4.509 x 10^2 mil/hr

vy ( velocity of airliner with respect to ground) =  -0.5877 x 10^2 mi/hr apprx

Magnitude = sqroot ( 4.509^2 + 0.5877^2 )X 10^2 = 4.547 x 10^ 2 mi/hr apprx

angle = tan^-1 ( 0.5877/  4.509) = 7.426 degree

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