Kinematics and Newtons 2nd law

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Homework Help Overview

The problem involves a baseball thrown vertically upward, requiring the application of Newton's second law to determine the ball's height as a function of time and its maximum height. The context is kinematics, specifically focusing on motion under the influence of gravity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Newton's second law but expresses uncertainty about deriving the height function. Some participants discuss the relationship between speed and time under gravitational acceleration, while others explore integration methods to find displacement.

Discussion Status

Participants are actively engaging with the problem, offering various approaches to derive the height function. There is a focus on initial conditions and constants in the integration process, with some guidance provided regarding the assignment of initial values. Multiple interpretations of the problem's requirements are being explored.

Contextual Notes

There is mention of the problem being relevant to a quantum chemistry course, indicating a potential overlap with calculus concepts. Participants are questioning the initial conditions and the appropriate calculus topics to review for this type of problem.

phillyj
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Homework Statement


A baseball of mass m is thrown vertically upward from a height r=0 with a speed of 20 meters/sec. The gravitational force on the baseball has a magnitude mg (m = mass, g=9.8 meters/sec2 is the acceleration due to gravity) and is directed downwards. Using Newton's second law, calculate the ball's height as a function of time and from that expression the maximum height of the ball.

Homework Equations



F=ma

The Attempt at a Solution



[tex]\sum{F=ma}[/tex][tex]\Rightarrow[/tex]mg=m[tex]\frac{d^{2}x}{dt^2}[/tex][tex]\Rightarrow[/tex]g=[tex]\frac{d^{2}x}{dt^2}[/tex]

That's the least I could think of and I'm not really sure. How exactly would I find the height from the second law?
 
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Since gravity is decreasing the ball's speed by 9.8 m/s every second and it starts with 20 m/s, how long does it take for the speed to reach 0?
 
Would that be 20/9.8=2.04 s?

I would like to know how to go about deriving the equation from the second law?
 
you can say that f / m = a and that a = 9.81, so f/m = 9.81
-9.81 = d^2x/dt^2
integrate both sides with respect to t
-9.81 * t + C(1) = dx/dt
integrate
-1/2 9.81 t^2 + c(1)t + c(2) = x(t)

It's an initial valued integral too so you can plug in the initial values to find the constants

I assume this is for a differential equations class since that approach is laborious and unused in most physics classes. Instead, people just use the results from the integration.
 
Last edited:
Yes, this is actually for quantum chemistry. Like a review/preview into the math we must know.

Its been a while since I did this so can you nudge me in the right direction? What are my initial values? I think that x(0), t is zero or am I thinking incorrectly.

Under what topic is calculus should I be reviewing for this sort of problem?
 
phillyj said:
Yes, this is actually for quantum chemistry. Like a review/preview into the math we must know.

Its been a while since I did this so can you nudge me in the right direction? What are my initial values? I think that x(0), t is zero or am I thinking incorrectly.

Under what topic is calculus should I be reviewing for this sort of problem?

c(1) is an arbitrary constant that must satisfy the initial conditions for a function of dx/dt(velocity!). At t = 0, we see that a * t = 0, so c(1) must account for the initial velocity alone. Thus, c(1) = v(initial) = 20 m/s

after further integration, we arrive at the function for x(t) (displacement) We see that at t = 0, 1/2 a t^2 + v(initial) t = 0, so c(2) must account for the initial position of the ball. In this problem, height begins at 0, so c(2) = 0.

your function for height x(t) becomes -1/2 9.81 t^2 + 20 t. The problem uses the variable r, so I'd switch it to r(t).

The negative sign denotes direction. I've assigned "toward earth" as negative and "toward sky" positive, so the initial upward velocity is positive and the constant acceleration is negative.
 

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