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 = [tex]ucos\theta*\frac{2usin\theta}{g} =\frac{u^2sin2\theta}{g}[/tex]
We have Range = 10 meter, θ = 18⁰
Substituting
[tex]10=\frac{u^2sin(2*18)}{9.8}\\ \\ u^2= 166.73\\ \\ u=12.91 m/s[/tex]
Takeoff speed of Kangaroo = 12.91 m/s
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
