Question

A ball is dropped from rest from a height of 200m. What is the height and velocity of the ball 3 seconds later?

Is this calc-based physics? If so, please walk me through it.

Answer 1

its not...

Answer 2

It's in my friends Calc 1 book, that's the only reason why I ask.

Answer 3

its easy just use these equatioons:
v=u+at;
s=ut+1/2at^2
v^2=u^2+2as

Answer 4

And u stands for....

Answer 5

initial velocity=u

Answer 6

@KenField

Answer 7

from rest means u=0

Answer 8

You have an expression for acceleration (g), integrate for velocity after a dt of 3 seconds, now you have v, integrate again for dx

Answer 9

to solve this in calc (though unnecessary, but ah well, the heck with it), we'll start with the following. acceleration is defined as:
\(\vec a = \frac{d \vec v }{d t}\)
rearraging, \( d \vec v = \vec a dt\)
integrating both sides with respect to t,
\(\vec v =\vec a t +C\)
since when \(t=0\), \(\vec v=0\), C=0.
but when \(t=3\), \(\vec a =-9.8 a_y\) where \(a_y\) is the unit vector in the y direction.
so, \(\vec v =-29.4a_y\)
now remember that \(\vec v =\frac{d \vec r}{dt}\)
integrating in the same way with acceleration,
\(\vec r = \vec v t +C\), where\( C=200a_y\) when t=0 and r=0.
now, substituting the values for \(\vec v_{t=3}\) and t=3,
\(\vec r = -9.8 a_y(3) +200a_y=170.6 a_y\)

is there a part where you need further help?

Answer 10

correction: \(\vec r=\vec v t+C\), where \(C=200a_y\) when t=0

Answer 11

Here's the easiest way in my opinion:
\[\large \text{distance fallen in 3 seconds}= \frac{1}{2}gt^2=\frac{1}{2}(-9.8m/s^2)(3.0s)=-44.1m\]\[\large \text{height after 3 seconds}=200m-44.1m=155.9m\]\[\large \text{velocity after 3 seconds}=at=(-9.8m/s^2)(3.0s)=-29.4m/s\]