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 February 19th, 2012, 01:32 PM #1 Newbie   Joined: Feb 2012 Posts: 5 Thanks: 0 Falling Object 1. If a ball is thrown vertically upward from the roof of a 32 foot building with a velocity of 32 ft/sec, then what is the maximum height the ball reaches? 2. What is the velocity of the ball when it hits the ground? Can someone please help me? I was never taught the formula or how to solve this kind of problem... And I'd be grateful to anyone that can help me xP And if you can... use the most simple words you can to explain this OTL Yeah, I'm dumb. Sorry
 February 19th, 2012, 02:07 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Falling Object There are two ways we can solve these questions: dynamics and conservation of energy. Dynamics: Once the ball is thrown, if we ignore drag, the only force working on the ball is that of gravity, so the acceleration of the ball is down with a magnitude of g. We can use the fact that acceleration is defined to be the time rate of change of velocity, and the given initial values to set up the initial value problem: $a=-g=\frac{dv}{dt}$ where $v(0)=v_0$ We also have the definition that velocity is the time rate of change of position, or $v=\frac{dx}{dt}$. We may remove time from the equation by using the chain rule as follows: $-g=\frac{dv}{dt}\cdot\frac{dt}{dx}\cdot v=v\cdot\frac{dv}{dx}$ where $x(0)=x_0$ Now, integrating with respect to x, using boundaries as limits, we find: $-g\int_{x_0}\,^x\,dx=\int_{v_0}\,^v v\,dv$ $-g$$x-x_0$$=\frac{1}{2}$$v^2-v_0^2$$$ $x=\frac{v_0^2-v^2}{2g}+x_0$ Now, when the object has reached its maximum height, its velocity is zero, thus we then have: $x=\frac{v_0^2}{2g}+x_0$ Now, it's just a matter of using the given data to find x. To find the velocity when the object hits the ground, set x to the value of the ground and solve for v: $x_{\text{ground}}=\frac{v_0^2-v^2}{2g}+x_0$ We tAke the negative root since the object is moving down: $v=-\sqrt{v_0^2+2g$$x_0-x_{\text{ground}}$$}$ Now, it's just a matter of using the given data to find v. I have to run now, but when I get back I will show how to use energy to solve these as well.
February 19th, 2012, 06:31 PM   #3
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Re: Falling Object

Hello, Ahsayuni!

Quote:
 1. If a ball is thrown vertically upward from the roof of a 32-foot building with a velocity of 32 ft/sec, what is the maximum height the ball reaches? 2. What is the velocity of the ball when it hits the ground?

$\text{In a="free fall=" problem, the height of the object is given by: }\:y(t) \:=\:h_o\,+\,v_ot\,-\,16t^2$

[color=beige]. . [/color]$\text{where: }\:\begin{Bmatrix}h_o &=& \text{initial height} \\ \\ \\
v_o &=& \text{initial velocity} \\ \\ \\ t &=& \text{time (in seconds)}\end{Bmatrix}$

$\text{W\!e are given: }\:h_o= 32,\;v_o = 32$

$\text{Our height function is: {\:y(t) \:=\:32\,+\,32t\,-\,16t^2$

$\text{The maximum height occurs where }y'(t) = 0.$

[color=beige]. . [/color]$y'(t) \,=\,32\,-\,32t \:=\:0 \;\;\;\Rightarrow\;\;\;t\,=\,1$

$\text{(1) The maximum height is: }\:y(1) \:=\:32\,+\,32(1)\,-\,16(1^2) \:=\:48\text{ feet.}$

$\text{="Hits the ground=" means }y(t) \.=\.0.$

[color=beige]. . [/color]$32\,+\,32t\,-\,16t^2\:=\:0 \;\;\;\Rightarrow\;\;\;t^2\,-\,2t\,-\,2 \:=\:0$

$\text{Quadratic Formula: }\:t \:=\:\frac{2\,\pm\sqrt{(-2)^2\,-\,4(1)(-2)}}{2(1)} \;=\; \frac{2\,\pm\,2\sqrt{3}}{2} \:=\:1\,\pm\,\sqrt{3}$

$\text{The ball hits the ground in }t \,=\,1 +\sqrt{3}\text{ seconds.}$

$\text{Velocity is: }\:v(t) \:=\:y#39;(t) \:=\:32\,-\,32t$

$(b)\;v(1+\sqrt{3}) \:=\:32\,-\,32(1\,+\,\sqrt{3}) \:=\:-32\sqrt{3} \;\approx\;-55.4\text{ ft/sec}$

 February 19th, 2012, 06:48 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Falling Object If we orient the axis of motion, the y-axis this time such that up is in the positive direction and the origin is at the ground, then the initial energy of the ball is part gravitational potential and part kinetic: $E_0=U_0+K_0$ $E_0=mgy_0+\frac{1}{2}mv_0^2$ When the ball reaches its maximum height, all of its energy is gravitational potential, but equal to the initial energy: $mgy_{\text{max}}=mgy_0+\frac{1}{2}mv_0^2$ $y_{\text{max}}=y_0+\frac{v_0^2}{2g}$ As you can see, this is the same result as in my first post. When the ball reaches the ground, all of its energy is kinetic energy, but equal to the initial energy: $\frac{1}{2}mv^2=mgy_0+\frac{1}{2}mv_0^2$ $v^2=2gy_0+v_0^2$ Again, taking the negative root since the ball is moving down: $v=-\sqrt{2gy_0+v_0^2}$ This is also equivalent to the result in my first post, given that $x_{\text{ground}}=0$.
 February 20th, 2012, 06:07 PM #5 Newbie   Joined: Feb 2012 Posts: 5 Thanks: 0 Re: Falling Object Thank you everyone that replied to my post! You've all been a great help @_____@ I finally understand how to do falling object problems now >__< Thanks~
 February 20th, 2012, 06:10 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Falling Object Glad to be a part of the help, and welcome to the forum!

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