Math Problem Statement

A kangaroo can jump over an object 2.09 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground. m/s (b) How long (in s) is it in the air? s

Solution

To solve this problem, we can use the basic equations of motion under constant acceleration, assuming the kangaroo's only acceleration is due to gravity (which is g=9.8m/s2g = 9.8 \, \text{m/s}^2).

Part (a) – Calculating the vertical speed when it leaves the ground

We know the following information:

  • Maximum height (hmaxh_{\text{max}}) = 2.09 m
  • Acceleration due to gravity (gg) = 9.8m/s29.8 \, \text{m/s}^2
  • At the maximum height, the vertical velocity is zero (v=0v = 0).

We can use the following kinematic equation to solve for the initial vertical speed (v0v_0):

v2=v022ghmaxv^2 = v_0^2 - 2g h_{\text{max}}

At the maximum height, v=0v = 0, so the equation simplifies to:

0=v022ghmax0 = v_0^2 - 2gh_{\text{max}}

Solving for v0v_0:

v0=2ghmaxv_0 = \sqrt{2gh_{\text{max}}}

Substitute the values:

v0=2×9.8×2.09v_0 = \sqrt{2 \times 9.8 \times 2.09}

Let's calculate that.

The vertical speed when the kangaroo leaves the ground is approximately 6.40m/s6.40 \, \text{m/s}.

Part (b) – Calculating the total time the kangaroo is in the air

To find the total time in the air, we can use the following kinematic equation:

t=2v0gt = \frac{2 v_0}{g}

This formula accounts for the time it takes to reach the maximum height and the time it takes to return to the ground (which are equal). Let's substitute the values and calculate the total time.

t=2×6.409.8t = \frac{2 \times 6.40}{9.8}

Let me compute that for you.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can still calculate the total time in the air using this formula:

t=2×6.409.8=1.31secondst = \frac{2 \times 6.40}{9.8} = 1.31 \, \text{seconds}

So the kangaroo is in the air for approximately 1.31 seconds.

Do you want more details or have any questions? Here are five related questions for further practice:

  1. How would the result change if the object was higher than 2.09 m?
  2. What would be the time of flight if the kangaroo jumped with a different initial velocity?
  3. How would air resistance affect the time in the air?
  4. Can this method apply to any vertical jump? Why or why not?
  5. How can we calculate the horizontal distance if the kangaroo jumped at an angle?

Tip: Always ensure that the units are consistent when working with physical equations (e.g., meters and seconds).

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Math Problem Analysis

Mathematical Concepts

Kinematics
Motion under gravity
Quadratic equations

Formulas

v_0 = sqrt(2gh_max)
t = 2v_0 / g

Theorems

Equations of motion under constant acceleration
Kinematic equations for vertical motion

Suitable Grade Level

Grades 9-12