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Physics EB5 A stone is thrown vertically up at 15.5 m/s from the edge of a cliff 75.0 m

karush

Well-known member
Jan 31, 2012
3,004
A stone is thrown vertically upward with a speed of 15.5 m/s
from the edge of a cliff 75.0 m high

ok, I was going to graph this with Desmos first
but don't know how to derive the equation for the parabola.

there are 11 questions following this...





$$\tiny{Embry-Riddle \, Aeronautical \, University, \, HW3 \, PS103}$$
 
Last edited:

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
A stone is thrown vertically upward with a speed of 15.5 m/s
from the edge of a cliff 75.0 m high

ok, I was going to graph this with Desmos first
but don't know how to derive the equation for the parabola.

there are 11 questions following this...
$f(t) = -\dfrac{1}{2}gt^{2} + v_{0}t + h_{0}$

$g$ is the Acceleration due to Gravity. Keep track of the sign and units.
$v_{0}$ is the Initial Velocity. Keep track of the sign and units.
$h_{0}$ is the Initial Height. Keep track of the sign and units.
$t$ is the time, in seconds.

Let's see your work.

P.S. Did I mention how important it is to keep track of the signs and units?
P.P.S. Burn this one into your head. You will need it.
 

karush

Well-known member
Jan 31, 2012
3,004
$f(t) = -\dfrac{1}{2}gt^{2} + v_{0}t + h_{0}$

$g$ is the Acceleration due to Gravity. Keep track of the sign and units.
$v_{0}$ is the Initial Velocity. Keep track of the sign and units.
$h_{0}$ is the Initial Height. Keep track of the sign and units.
$t$ is the time, in seconds.
$\begin{align*}
f(t)&=-\dfrac{1}{2}gt^{2}+v_{0}t+h_{0}\\
&=-0.5(9.8)t^2+15.5t+75
\end{align*}$
ok I wasn't sure sure about $g$ but so far this
eb5.png
a. How much time will it take to reach the highest point?
$\begin{align*}
s(t)&=-\dfrac{1}{2}gt^{2}+v_{0}t+h_{0}\\
v(t)&=-9.8t+15.5 \\
v(1.58163)&=0
\end{align*}$
well I got $\approx 1.582s$
 
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tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
Looks good.

It is important to remember that this model refers to the VERTICAL displacement ONLY. It is easy to mistake the pretty picture to mean something about horizontal displacement. We threw the object STRAIGHT UP and it falls STRAIGHT DOWN. Nothing horizontal in this model.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
That is, the horizontal axis of your graph is time, not distance.
 

karush

Well-known member
Jan 31, 2012
3,004
Looks good.

It is important to remember that this model refers to the VERTICAL displacement ONLY. It is easy to mistake the pretty picture to mean something about horizontal displacement. We threw the object STRAIGHT UP and it falls STRAIGHT DOWN. Nothing horizontal in this model.
however we are not graphing the actual path the object
but rhe speed of the objecr which is a parabola !
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
however we are not graphing the actual path the object
but rhe speed of the objecr which is a parabola !
Speed? No. This is just vertical displacement - position.

You correctly observed where the velocity was zero (0) by finding the first derivative.

If you throw the stone in view of a fixed camera position, you would see no parabola - Just up and down.
If you repositioned the camera consistently and quickly in one direction, then you would see the parabola.

Vertical axis is vertical displacement (height).
Horizontal axis is time.
 

karush

Well-known member
Jan 31, 2012
3,004
Vertical axis is vertical displacement (height).
Horizontal axis is time.
If it was not a parabola the t-axis
Would useless.
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
If it was not a parabola the t-axis
Would useless.
I would tend to disagree and reverse your argument. Without the t-axis, it isn't a parabola. There isn't anything about throwing a rock straight up that says "parabola" UNLESS you watch the thing as it travels through time.