A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 11 from the takeoff point.

If the kangaroo leaves the ground at a 18 angle, what is its takeoff speed ?
What is its horizontal speed?

Well, let me hop right into the answer for you! If the gray kangaroo is jumping with an angle of 18 degrees and covering a distance of 11 meters in each jump, we can use some math-tricks to calculate its takeoff speed and horizontal speed.

To calculate the takeoff speed, we can use some trigonometry. The takeoff speed can be found using the equation: takeoff speed = horizontal speed / cos(angle). So, if the kangaroo is covering 11 meters in each jump and the angle is 18 degrees, we can calculate the takeoff speed as follows:

takeoff speed = 11 meters / cos(18 degrees)

Now, to calculate the horizontal speed, we can use the equation: horizontal speed = takeoff speed * cos(angle). Again, using the given values:

horizontal speed = takeoff speed * cos(18 degrees)

And voila, you've got your answer! Just substitute the values and calculate away. Remember, clowns are notoriously bad at math, but I'm here to bring some laughter into the equation!

To find the takeoff speed of the gray kangaroo, we need to use trigonometry.

We are given that the kangaroo jumps 11m from the takeoff point and leaves the ground at an angle of 18 degrees.

First, let's find the vertical component of the takeoff speed. We can use the equation:

Vertical component = Takeoff speed * sin(angle)

Vertical component = Takeoff speed * sin(18)

Next, let's find the horizontal component of the takeoff speed. We can use the equation:

Horizontal component = Takeoff speed * cos(angle)

Horizontal component = Takeoff speed * cos(18)

Since the kangaroo's initial velocity is directly upwards, the vertical component of the takeoff speed is equal to the initial velocity of the kangaroo. Therefore, we can write:

Vertical component = Initial velocity = 11

Now, we can substitute the known values into the equations to solve for the takeoff speed and the horizontal speed.

Vertical component = Takeoff speed * sin(18)

11 = Takeoff speed * sin(18)

Takeoff speed = 11 / sin(18)

Takeoff speed ≈ 38.44 m/s

Horizontal component = Takeoff speed * cos(18)

Horizontal speed = 38.44 * cos(18)

Horizontal speed ≈ 36.95 m/s

Therefore, the takeoff speed of the kangaroo is approximately 38.44 m/s, and its horizontal speed is approximately 36.95 m/s.

To find the takeoff speed and horizontal speed of the kangaroo, we can break down the problem into its horizontal and vertical components.

First, let's find the vertical component of the kangaroo's initial velocity.
The vertical component of the takeoff speed can be found using the formula:
Vertical Speed = Takeoff Speed * sin(θ),
where θ is the angle of takeoff, and sin(θ) represents the sine of the angle.

In this case, the angle of takeoff is 18 degrees.
So, the vertical component of the takeoff speed can be calculated as follows:
Vertical Speed = Takeoff Speed * sin(18)

Next, let's find the horizontal component of the kangaroo's initial velocity.
The horizontal component of the takeoff speed can be found using the formula:
Horizontal Speed = Takeoff Speed * cos(θ),
where θ is the angle of takeoff, and cos(θ) represents the cosine of the angle.

In this case, the angle of takeoff is 18 degrees.
So, the horizontal component of the takeoff speed can be calculated as follows:
Horizontal Speed = Takeoff Speed * cos(18)

Now let's solve for the takeoff speed. We know that the kangaroo covers a distance of 11 meters with each jump. This distance represents the total displacement of the kangaroo, which is the combination of the vertical and horizontal components.

Using the Pythagorean theorem, we can express the relationship between the vertical and horizontal components of the initial velocity and the total displacement:
Vertical Speed * time = -11 (since the initial vertical velocity is opposite to the direction of gravity, we use -11 instead of 11)
Horizontal Speed * time = 11

Here, time represents the total time taken for the kangaroo to cover the horizontal distance of 11 meters.

By dividing these two equations, we can eliminate the variable time and solve for the takeoff speed:
(Takeoff Speed * sin(18)) / (Takeoff Speed * cos(18)) = -11 / 11

Simplifying the equation yields:
tan(18) = -1 (since sin(θ) / cos(θ) = tan(θ))

Solving for tan(18), we find:
tan(18) ≈ 0.3249

Therefore, the takeoff speed can be calculated as:
Takeoff Speed = -11 / tan(18)

To find the horizontal speed, substitute the value of the takeoff speed into the equation:
Horizontal Speed = Takeoff Speed * cos(18)

Evaluating this equation, we can find the horizontal speed of the kangaroo.

Takeoff Speed= Range=(initial velocity^2 * Sin (2 theta))/g

11m = (V0^2*sin(36))/(9.8)

V0= sqrt((11*9.8)/(sin36))

V0=13.54