Kinematics and Newtons 2nd law

AI Thread Summary
The discussion focuses on calculating the height of a baseball thrown vertically upward using Newton's second law. The gravitational force acting on the baseball is expressed as mg, where g is 9.8 m/s². The participants derive the height function x(t) through integration, leading to the equation x(t) = -1/2 * 9.81 * t² + 20 * t, accounting for initial conditions of velocity and position. The negative sign in the equation indicates the downward direction of gravitational acceleration. The conversation highlights the importance of understanding initial values and the integration process in solving kinematic problems.
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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



\sum{F=ma}\Rightarrowmg=m\frac{d^{2}x}{dt^2}\Rightarrowg=\frac{d^{2}x}{dt^2}

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.
 
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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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