<span>Each jump = 11 m
Angle of the speed = 21 degrees.
Gravitational Acceleration g = 9.8 m/s^2 Initial velocity V^2 = (Rg)/sin2A
V^2 = 11 x 9.81 / sin (2 x 21)
V^2 = 107. 91 / 0.669
V^2 = 161.27
V = 12.70
Now we got the velocity and calculate the horizontal angle,
Velocity (horizontal) = vcosA = 12.70 x cos 21
Velocity = 11.85 m/s</span>
We have that the maximum rank of the kangaroo is given by: R = v0 ^ 2 sin (2θ) / g where, v0 = initial velocity θ = angle of the velocity vector formed from the horizontal g = gravity Clearing the speed we have: v0 ^ 2 = (R * g) / (sin (2θ)) Substituting values v0 = root (((11) * (9.8)) / (sin (2 (21 * (pi / 180))))) v0 = 12.69 m / s answer its takeoff speed is 12.69 m / s
I think it will increase a little bit ... just image ... if the temperature is 0, the velocity will be 0 too. because the vibration of atom is so weak and the sound cant progation.