A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 9.3 m from the takeoff point. If the kangaroo leaves the ground at a 17.6 degree angle, what is its takeoff speed?
The distance covered in each jump is
9.3 m = (Vo^2/g)*sin(2*17.6)
g = 9.8 m/s^2
9.3 = [Vo^2/9.8]*0.5764
Solve for Vo
To find the takeoff speed of the gray kangaroo, we can use the principles of projectile motion. Projectile motion involves the motion of an object in two dimensions under the influence of only gravity and no other forces.
In this case, we have a kangaroo jumping in a flat stretch, which means the motion can be treated as a projectile motion.
The first step is to split the initial velocity into horizontal and vertical components. The horizontal component of velocity remains constant throughout the motion, while the vertical component is affected by gravity.
Let's assume the takeoff speed as "V". We can find the initial horizontal velocity (Vx) and initial vertical velocity (Vy) using trigonometry:
Vx = V * cos(θ)
Vy = V * sin(θ)
Where θ is the angle at which the kangaroo leaves the ground (17.6 degrees in this case).
Since the horizontal component of velocity remains constant, we can use the horizontal displacement formula:
Horizontal displacement = Vx * Time
In this case, the horizontal displacement is given as 9.3 meters, so we can rearrange the formula to solve for time:
Time = Horizontal displacement / Vx
Once we have the time, we can use it to find the vertical displacement using the kinematic equation:
Vertical displacement = Vy * Time + (0.5) * (-9.8) * (Time)^2
In this equation, (0.5) * (-9.8) is the acceleration due to gravity, and it acts in the downward direction.
Since the kangaroo lands at the same height as it started, the vertical displacement will be zero:
0 = Vy * Time + (0.5) * (-9.8) * (Time)^2
Now, we can substitute the values of Vx, Vy, and Time in terms of V and solve this equation to find the takeoff speed (V).