Math Problem Statement

A frog jumps into the air, leaving the ground at an angle of 30˚ above the horizontal. The frog reaches a maximum height of 0.6 m. a) With what velocity does the frog leave the ground? b) How long is the frog in the air from takeoff until landing?

Solution

To solve this problem, we use kinematic equations for projectile motion. We'll break the initial velocity (v0v_0) into horizontal and vertical components, analyze the vertical motion for maximum height and time of flight, and then use this to answer the given questions.


Given:

  • Launch angle (θ\theta) = 3030^\circ
  • Maximum height (hmaxh_{\text{max}}) = 0.6m0.6 \, \text{m}
  • Acceleration due to gravity (gg) = 9.8m/s29.8 \, \text{m/s}^2

Step 1: Find the vertical velocity (vy0v_{y0}) at launch

At the maximum height, the vertical velocity becomes zero. Using the kinematic equation: vy2=vy022ghmaxv_y^2 = v_{y0}^2 - 2 g h_{\text{max}} At maximum height, vy=0v_y = 0. Thus: 0=vy022ghmax0 = v_{y0}^2 - 2 g h_{\text{max}} vy0=2ghmaxv_{y0} = \sqrt{2 g h_{\text{max}}} Substitute the known values: vy0=29.80.6=11.763.43m/sv_{y0} = \sqrt{2 \cdot 9.8 \cdot 0.6} = \sqrt{11.76} \approx 3.43 \, \text{m/s}


Step 2: Find the initial velocity (v0v_0) at launch

The vertical component of velocity is related to the total initial velocity by: vy0=v0sinθv_{y0} = v_0 \sin\theta Rearrange to solve for v0v_0: v0=vy0sinθv_0 = \frac{v_{y0}}{\sin\theta} Substitute vy0=3.43m/sv_{y0} = 3.43 \, \text{m/s} and sin(30)=0.5\sin(30^\circ) = 0.5: v0=3.430.56.86m/sv_0 = \frac{3.43}{0.5} \approx 6.86 \, \text{m/s}


Step 3: Find the total time of flight (TT)

The total time of flight is the time it takes for the frog to go up and come down. The time to reach maximum height (tupt_{\text{up}}) is given by: tup=vy0gt_{\text{up}} = \frac{v_{y0}}{g} Substitute vy0=3.43m/sv_{y0} = 3.43 \, \text{m/s} and g=9.8m/s2g = 9.8 \, \text{m/s}^2: tup=3.439.80.35st_{\text{up}} = \frac{3.43}{9.8} \approx 0.35 \, \text{s} The total time of flight is twice this value: T=2tup=20.350.70sT = 2 t_{\text{up}} = 2 \cdot 0.35 \approx 0.70 \, \text{s}


Final Answers:

(a) The velocity with which the frog leaves the ground is approximately: 6.86m/s\boxed{6.86 \, \text{m/s}} (b) The total time the frog is in the air is approximately: 0.70s\boxed{0.70 \, \text{s}}


Would you like detailed step-by-step derivations for any part?

Here are 5 related questions for deeper understanding:

  1. How can you calculate the range of the frog's jump?
  2. How does the launch angle affect the maximum height?
  3. What happens to the time of flight if the frog jumps with a steeper angle but the same speed?
  4. How would the calculations change if the ground was not level (e.g., landing on a hill)?
  5. Can you calculate the horizontal distance covered by the frog?

Tip: Always separate the motion into horizontal and vertical components when solving projectile problems!

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematics
Trigonometry

Formulas

v_y^2 = v_{y0}^2 - 2gh
v_{y0} = v_0 sin(θ)
t_{up} = v_{y0} / g
Total time of flight T = 2t_{up}

Theorems

Kinematic equations of motion
Trigonometric relationships in projectile motion

Suitable Grade Level

Grades 9-12