Question

A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 11 m from the takeoff point. part a if the kangaroo leaves the ground at a 21 ∘ angle, what is its takeoff speed

Answer 1

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

Answer 2

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