Math Problem Statement

A football is punted from the ground, it passes through the apex of its flight, and it begins its downward descent, as shown. When its height above the field is h = 1.6 m, the horizontal and vertical components of its velocity are observed to have the values v_horizontal = 18.5 m/s and v_vertical = 2.5 m/s, respectively. Determine the magnitude of the horizontal component of the football's initial velocity in terms of v_horizontal, v_vertical, g, and h, and calculate the time from when the football left the ground until it is descending at a height of h = 1.6 m.

Solution

The problem provided involves analyzing the motion of a football in projectile motion. It contains two parts:

  • Part (a): Determine the magnitude of the horizontal component of the football's initial velocity, v0,xv_{0,x}, in terms of the horizontal and vertical velocities (vhorizontalv_{\text{horizontal}}, vverticalv_{\text{vertical}}), gravity (gg), and height (hh).

  • Part (b): Calculate the time from when the football left the ground until it is descending at a height of h=1.6mh = 1.6 \, \text{m}.

Let's address Part (a) first.

Part (a) Explanation:

The horizontal velocity of a projectile remains constant throughout its flight, assuming no air resistance. Given that the horizontal velocity at a height of h=1.6mh = 1.6 \, \text{m} is already provided as vhorizontal=18.5m/sv_{\text{horizontal}} = 18.5 \, \text{m/s}, this value is equal to the initial horizontal velocity:

v0,x=vhorizontal=18.5m/s.v_{0,x} = v_{\text{horizontal}} = 18.5 \, \text{m/s}.

Part (b) Solution:

To solve for the time, we use the vertical motion equation. The vertical motion is affected by gravity and can be described by the kinematic equation:

vy2=v0,y22g(hh0),v_{y}^2 = v_{0,y}^2 - 2g(h - h_0),

where:

  • vy=2.5m/sv_{y} = -2.5 \, \text{m/s} (the vertical velocity when descending at h=1.6mh = 1.6 \, \text{m}),
  • v0,yv_{0,y} is the initial vertical velocity (unknown for now),
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2,
  • h=1.6mh = 1.6 \, \text{m} (the height given),
  • h0=0h_0 = 0 (since the football starts from the ground).

We can first find v0,yv_{0,y} and then use the vertical motion equations to find the time. Let me calculate that for you.The initial vertical velocity v0,yv_{0,y} is approximately 6.13m/s6.13 \, \text{m/s}.

Next, to find the total time from when the football was kicked until it reaches a height of 1.6m1.6 \, \text{m} while descending, we use the following kinematic equation for vertical motion:

vy=v0,ygt.v_y = v_{0,y} - g \cdot t.

Solving for tt, the time:

t=v0,yvyg.t = \frac{v_{0,y} - v_y}{g}.

Let's calculate the time now.It seems like I can't do more advanced data analysis right now. Please try again later.

However, you can still calculate the time using the formula:

t=v0,yvyg,t = \frac{v_{0,y} - v_y}{g},

where v0,y6.13m/sv_{0,y} \approx 6.13 \, \text{m/s}, vy=2.5m/sv_y = -2.5 \, \text{m/s}, and g=9.8m/s2g = 9.8 \, \text{m/s}^2.

If you need further assistance or have any other questions, feel free to ask!

Here are 5 related questions to think about:

  1. How would air resistance affect the football's motion?
  2. What is the total time of flight for the football?
  3. How can we determine the maximum height the football reaches?
  4. What is the football's range (horizontal distance covered)?
  5. How would the problem change if the football was kicked at an angle different from 45°?

Tip: The horizontal and vertical motions of a projectile are independent of each other. This can simplify the way you approach problems related to projectile motion.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematic Equations
Two-Dimensional Motion

Formulas

v_0x = v_horizontal
v_y^2 = v_0y^2 - 2g(h - h_0)
t = (v_0y - v_y) / g

Theorems

Kinematic Equation for Vertical Motion
Independence of Horizontal and Vertical Components of Motion

Suitable Grade Level

Grades 11-12