Agray 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

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

It may possibly be c i think |-/