Question

A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 10 m from the takeoff point. you may want to review ( pages 83 - 85) . part a if the kangaroo leaves the ground at a 18 â angle, what is its takeoff speed ?

Answer 1

Answer:

Takeoff speed of Kangaroo = 12.91 m/s

Explanation:

  The distance reached by Kangaroo =  10 meter

  Angle at which it jumps = 18°

  The motion of Kangaroo is like a projectile, the distance traveled is the range of projectile.

   Range of projectile = Time taken for the projectile to reach ground* Horizontal velocity

   Time taken for the projectile to reach ground:

       Time taken = Two times of the time taken for the projectile to reach maximum height

       Time taken for the projectile to reach maximum height = Vertical speed / Acceleration = u sin θ/g

      Time taken for the projectile to reach ground = 2 u sin θ/g

  So Range of projectile = $ucos\theta*\frac{2usin\theta}{g} =\frac{u^2sin2\theta}{g}$

 We have Range = 10 meter, θ = 18⁰

    Substituting

        $10=\frac{u^2sin(2*18)}{9.8}\\ \\ u^2= 166.73\\ \\ u=12.91 m/s$

 Takeoff speed of Kangaroo = 12.91 m/s

Answer 2

Answer to Part B of this question (what is the horizontal speed?):

Answer (in meters/second):

(takeoff speed) * cos(angle given... 18 for this problem) = horizontal speed

Explanation:

For anyone looking to find the answer to part B of this problem on Mastering Physics (this worked for my homework):

Part B: What is its horizontal speed?

Answer: (takeoff speed) * cos(angle given... 18 for this problem) = horizontal speed