A gray kangaroo can bound across level ground with each jump carrying it 8.4 from the takeoff point. Typically the kangaroo leaves the ground at a 23 angle. What is its takeoff speed? What is its maximum height above the ground?

To determine the takeoff speed of a kangaroo, we can use the horizontal and vertical components of its motion.

1. Takeoff speed:
The horizontal component of the kangaroo's motion remains constant throughout the jump, while the vertical component is affected by gravity. We can use the horizontal motion to find the takeoff speed.

In this case, the kangaroo covers a horizontal distance of 8.4 meters with an angle of 23 degrees from the takeoff point. To find the horizontal component (Vx) of the takeoff speed, we can use trigonometric ratios.

Vx = velocity * cos(angle)

Given that the horizontal distance is 8.4 meters and the angle of takeoff is 23 degrees, we can calculate the horizontal component:

Vx = 8.4 * cos(23°)

Using a scientific calculator, we find:

Vx ≈ 7.79 m/s

Therefore, the takeoff speed of the kangaroo is approximately 7.79 m/s horizontally.

2. Maximum height:
To find the maximum height above the ground, we need to calculate the vertical component (Vy) of the takeoff speed.

Vy = velocity * sin(angle)

Using the same angle of takeoff (23 degrees), we can calculate the vertical component:

Vy = 7.79 * sin(23°)

Using a scientific calculator, we find:

Vy ≈ 3.19 m/s

Now, we can find the time it takes for the kangaroo to reach its maximum height, ignoring air resistance. We know that at the maximum height, the vertical component of the velocity becomes zero.

Using the formula for vertical motion under constant acceleration:

Vy = Voy + (-g * t)
0 = 3.19 - (9.8 * t)

Solving for t:

t ≈ 0.326 seconds

Finally, we can find the maximum height (H) using the formula:

H = Voy * t + (1/2) * (-g) * t^2

H = 3.19 * 0.326 + (1/2) * -9.8 * (0.326^2)

Using a scientific calculator, we find:

H ≈ 1.167 meters

Therefore, the gray kangaroo reaches a maximum height of approximately 1.167 meters above the ground.

speed*cos23*timeinair=8.4

speed= you solve it in terms of t.

then vertical equation:

hf=hi-4.9t^2+speed*sin23*t

put what you solved for speed above into this equation, then solve for timeinair.
then solve for speed.