# Story drop Problems

• Nov 9th 2006, 04:06 PM
killasnake
Story drop Problems
How do I solve these?

Quote:

A stone is dropped from the edge of a roof, and hits the ground with a velocity of -165 feet per second. How high (in feet) is the roof?
and

Quote:

A particle is moving with acceleration a(t) = 24t+4. Its position at time t=0 is s(0) = 14 and its velocity at time t=0 is v(0) = 3 What is its position at time t=6? ________
• Nov 9th 2006, 04:20 PM
Quick
Quote:

Originally Posted by killasnake
A stone is dropped from the edge of a roof, and hits the ground with a velocity of -165 feet per second. How high (in feet) is the roof?

$v_i=0$ because it was from rest

$v_f=-165$ I suppose down is negative in this problem

$a=-9.8$ If you don't know why then you're in trouble, however that is in m/s so you have to convert to feet (which I forgot the conversion so you'll have to do it yourself)

$\Delta d=?$ This is what we're trying to find

Notice that they don't give time, so we'll use an equation without time.

The only one that comes to mind is: $v_f^2=v_i^2+2a\Delta d$

So then we switch things around to get: $\Delta d=\frac{v_f^2-v_i^2}{2a}$

So substitute to get the answer :)
• Nov 9th 2006, 04:23 PM
topsquark
Quote:

Originally Posted by Quick
$v_i=0$ because it was from rest

$v_f=-165$ I suppose down is negative in this problem

$a=-9.8$ If you don't know why then you're in trouble, however that is in m/s so you have to convert to feet (which I forgot the conversion so you'll have to do it yourself)

$\Delta d=?$ This is what we're trying to find

Notice that they don't give time, so we'll use an equation without time.

The only one that comes to mind is: $v_f^2=v_i^2+2a\Delta d$

So then we switch things around to get: $\Delta d=\frac{v_f^2-v_i^2}{2a}$

So substitute to get the answer :)

That does fairly well for an answer Quick, except for one little problem:
Acceleration is in units of m/s^2, not m/s. (Or ft/s^2 if you prefer.)

Tsk, tsk! This is the second time I've had to tell you that!

-Dan
• Nov 9th 2006, 04:32 PM
AfterShock
Quote:

Originally Posted by Quick

$a=-9.8$ If you don't know why then you're in trouble, however that is in m/s so you have to convert to feet (which I forgot the conversion so you'll have to do it yourself)

s(t) = -16t^2 ft
v(t) = -32t ft/s
a(t) = -32 ft/s^2
• Nov 9th 2006, 04:32 PM
topsquark
Quote:

Originally Posted by killasnake
A particle is moving with acceleration a(t) = 24t+4. Its position at time t=0 is s(0) = 14 and its velocity at time t=0 is v(0) = 3 What is its position at time t=6?

We've got the acceleration function, and we know s(0) and v(0) as initial conditions. So:
$v(t) - v(0) = \int_0^t \, a(t') \, dt' = \int_0^t \, (24t' + 4)dt' = (12t'^2 + 4t')|_0^t = 12t^2 + 4t$

So $v(t) = 12t^2 + 4t - 3$

Again:
$s(t) - s(0) = \int_0^t \, v(t') \, dt' = \int_0^t \, (12t'^2 + 4t' - 3)dt'$ $= (4t'^3 + 2t'^2 - 3t')|_0^t = 4t^3 + 2t^2 - 3t$

So $s(t) = 4t^3 + 2t^2 - 3t - 14$

Thus $s(6) = 4(6)^3 + 2(6)^2 - 3(6) - 14 = 904$ (in whatever units the problem is supposed to be written in.)

Piece of trivia for you: The coefficient of the t^3 term is constant. This means that the above equation for displacement is written in terms of a constant "jerk." (No kidding! The jerk, j, is defined as the first time derivative of the acceleration.)

-Dan
• Nov 9th 2006, 04:56 PM
killasnake
:eek: wow you guys are quick! thank you so much for the help!
• Nov 9th 2006, 06:22 PM
Soroban
Hello, killasnake!

Quote:

1) A stone is dropped from the edge of a roof,
and hits the ground with a velocity of -165 ft/sec.
How high (in feet) is the roof?

You're expected to know the "free fall" formula: . $y(t) \:=\:h_o + v_ot - 16t^2$

. . where: . $\begin{Bmatrix}y(t) = & \text{height at time }t \\ h_o = & \text{initial height} \\ v_o = & \text{initial velocity} \\ t = & \text{time in seconds}\end{Bmatrix}$

In this problem, the stone is dropped: $v_o = 0$
. . Hence, the function is: . $y(t) \:=\:h_o - 16t^2$
and we want to determine $h_o.$

We are told that the stone hits the ground at -165 ft/sec.
When does this happen?
. . It happens when $y(t) = 0$
So we have: . $h_o - 16t^2\:=\:0\quad\Rightarrow\quad t = \frac{\sqrt{h_o}}{4}$ seconds.

Velocity is the derivative of position: . $v(t)\:=\:-32t$

At that time, we have: . $-32t \:=\:-165\quad\Rightarrow\quad-32\left(\frac{\sqrt{h_o}}{4}\right)\:=\:-165$

. . $-8\sqrt{h_o} \:=\:-165\quad\Rightarrow\quad \sqrt{h_o}\:=\:\frac{165}{8}\quad\Rightarrow\quad h_o \:=\:\left(\frac{165}{8}\right)^2$

Therefore: . $h_o\:=\:425.390625 \:\approx\:425.4$ feet.

• Nov 13th 2006, 05:11 PM
thedoge
Quote:

Originally Posted by topsquark
We've got the acceleration function, and we know s(0) and v(0) as initial conditions. So:
$v(t) - v(0) = \int_0^t \, a(t') \, dt' = \int_0^t \, (24t' + 4)dt' = (12t'^2 + 4t')|_0^t = 12t^2 + 4t$

So $v(t) = 12t^2 + 4t - 3$

Again:
$s(t) - s(0) = \int_0^t \, v(t') \, dt' = \int_0^t \, (12t'^2 + 4t' - 3)dt'$ $= (4t'^3 + 2t'^2 - 3t')|_0^t = 4t^3 + 2t^2 - 3t$

So $s(t) = 4t^3 + 2t^2 - 3t - 14$

Thus $s(6) = 4(6)^3 + 2(6)^2 - 3(6) - 14 = 904$ (in whatever units the problem is supposed to be written in.)
-Dan

Slight correction on this problem.

v(t) is NOT $v(t) = 12t^2 + 4t - 3$
it is $v(t) = 12t^2 + 4t + 3$

and hence

s(t) is NOT $s(t) = 4t^3 + 2t^2 - 3t - 14$
it is $s(t) = 4t^3 + 2t^2 + 3t + 14$

so the answer is 968 not 904