# Homework Help: Kinematics and Newtons 2nd law

1. Jan 24, 2010

### phillyj

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

F=ma

3. The attempt at a solution

$$\sum{F=ma}$$$$\Rightarrow$$mg=m$$\frac{d^{2}x}{dt^2}$$$$\Rightarrow$$g=$$\frac{d^{2}x}{dt^2}$$

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?

2. Jan 24, 2010

### ideasrule

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?

3. Jan 24, 2010

### phillyj

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?

4. Jan 24, 2010

### xcvxcvvc

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: Jan 24, 2010
5. Jan 24, 2010

### phillyj

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?

6. Jan 24, 2010

### xcvxcvvc

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.