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

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

Answer 2

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