a scared kangaroo clears a fence that is 2.44 meters tall. what is the vertical component of the kangaroos velocity at take off (vfi) ? if the horizontal component of the kangaroos velocity is 4.80m/s at what angle did the kangaroo take off the ground? (hint: at its maximum height the kangaroos vertical velocity is 0)
Vertical component is 6.915 m
To find the vertical component of the kangaroo's velocity at takeoff, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad,
where vf is the final vertical velocity (which is zero since at its maximum height the kangaroo's vertical velocity is 0), vi is the initial vertical velocity, a is the acceleration (which is approximately -9.8 m/s^2 due to gravity), and d is the displacement (which is the height of the fence, 2.44 meters).
Since vf^2 = vi^2 + 2ad, and vf = 0, we can rearrange the equation to solve for vi:
0 = vi^2 + 2(-9.8)(2.44).
Simplifying this equation gives us:
0 = vi^2 - 47.872.
Rearranging the equation again gives us:
vi^2 = 47.872.
Taking the square root of both sides gives us:
vi = √(47.872).
Therefore, the vertical component of the kangaroo's velocity at takeoff (vfi) is approximately √(47.872) m/s.
To find the angle at which the kangaroo takes off, we can use the concept of vectors. The horizontal component of velocity (Vx) can be given by:
Vx = vi * cosθ,
where vi is the initial velocity and θ is the angle of takeoff.
Since Vx = 4.80 m/s, and vi = √(47.872) m/s (as found earlier), we can substitute these values into the equation:
4.80 = √(47.872) * cosθ.
To solve for θ, we can rearrange the equation:
cosθ = 4.80 / √(47.872).
Taking the inverse cosine (cos^-1) of both sides gives us:
θ = cos^-1(4.80 / √(47.872)).
Evaluating this expression will give us the angle at which the kangaroo took off from the ground.
To calculate the vertical component of the kangaroo's velocity at takeoff (Vfi), we will need to use the concept of projectile motion. We can break down the motion into vertical and horizontal components.
1. Finding the vertical component of velocity (Vfi):
- We know that the maximum height reached by the kangaroo is when its vertical velocity (Vy) becomes zero.
- We can use the kinematic equation: Vy = Voy + gt, where Voy is the initial vertical velocity and g is the acceleration due to gravity (-9.8 m/s²).
- At the maximum height, Vy = 0, so the equation becomes 0 = Voy + (-9.8)t.
- We know that the time it takes for the kangaroo to reach its maximum height is the same as the time it takes to fall back down from the top of the parabolic trajectory.
- Therefore, we can divide the total time in half to get the time at the maximum height, t/2.
- Solving the equation, 0 = Voy + (-9.8)(t/2).
- Rearranging the equation, we can find the initial vertical velocity Voy: Voy = 9.8(t/2).
Let's calculate the time it takes for the kangaroo to reach the maximum height (t):
- We can use the equation s = ut + (1/2)at², where s is the displacement, u is the initial velocity, a is acceleration, and t is time.
- In this case, s = 2.44 m (height of the fence), u = Voy, a = g, and we need to solve for t.
- The displacement is the height of the fence, which is equal to the displacement when the kangaroo starts from the ground.
- So, 2.44 m = 0 + (1/2)(-9.8)t².
- Simplifying the equation, 4.9t² = 2.44.
- Dividing both sides by 4.9, we get t² = 2.44/4.9.
- Taking the square root of both sides, t = sqrt(2.44/4.9).
Now we can substitute the value of t/2 into the equation and find the initial vertical velocity (Voy):
- Voy = 9.8(t/2) = 9.8(sqrt(2.44/4.9)/2).
2. Finding the angle at which the kangaroo takes off:
- We know the horizontal component of velocity (Vx) is 4.80 m/s.
- The vertical component of velocity (Vy) at takeoff is the same as the component of velocity when the kangaroo is at its maximum height (Vfi).
- We can calculate the angle using the formula: tan(θ) = Vy / Vx, where θ is the angle.
- Rearranging the equation, we get θ = arctan(Vy / Vx).
Now we can substitute the calculated values of Vy and Vx into the equation and find the angle (θ):
- θ = arctan(Voy / Vx) = arctan(9.8(sqrt(2.44/4.9)/2) / 4.80).
By substituting the values and evaluating the equation, you can find the vertical component of velocity (Vfi) and the angle (θ) at which the kangaroo takes off.
Vfi^2 = 2gH
H = 2.44 m, g = 9.8 m
Solve for Vfi, the initial vertical component.
Vfi/4.80 = tanA
Solve for the angle A.