# Thread: motion of a projectile

1. ## motion of a projectile

Here's a doozy for ya.
A projectile is fired directly upward with a muzzle velocity of 860 feet per second from a height of 7 feet above the ground.

A) Determine a function for the height of the projectile t seconds after it's released.
B) How long does it take the projectile to reach a height of 100 feet on its way up?
C) How long is the projectile in the air?

So far, this is what I have for A.)
X=(860cos90*) + and then I get lost. Any takers????

2. Originally Posted by suzdisp
Here's a doozy for ya.
A projectile is fired directly upward with a muzzle velocity of 860 feet per second from a height of 7 feet above the ground.

A) Determine a function for the height of the projectile t seconds after it's released.
B) How long does it take the projectile to reach a height of 100 feet on its way up?
C) How long is the projectile in the air?

So far, this is what I have for A.)
X=(860cos90*) + and then I get lost. Any takers????
The particle is fired directly upward, so we don't need to worry about x components or any angles. Thus the form for the height of the projectile is given by
$\displaystyle h = h_0 + v_0t + \frac{1}{2}at^2$
where $\displaystyle h_0$ is the height at t = 0, $\displaystyle v_0$ is the initial velocity, and a is the acceleration which in this case is -g = -32 ft/s^2.

$\displaystyle h = 7 + 860t - 16t^2$

See what you can do from there.

-Dan

3. Originally Posted by suzdisp
Here's a doozy for ya.
A projectile is fired directly upward with a muzzle velocity of 860 feet per second from a height of 7 feet above the ground.

A) Determine a function for the height of the projectile t seconds after it's released.
B) How long does it take the projectile to reach a height of 100 feet on its way up?
C) How long is the projectile in the air?

So far, this is what I have for A.)
X=(860cos90*) + and then I get lost. Any takers????
$\displaystyle g=-\frac{32ft}{s^2} \mbox{ and } v_0=\frac{860ft}{2} \mbox{ and }x_0=7$

Using the kinematic equation $\displaystyle x=\frac{1}{2}gt^2+v_0t+x_0$

gives us $\displaystyle x=-16t^2+860t+7$

for b solve $\displaystyle 100=-16t^2+860t+7$

for c solve $\displaystyle 0=-16t^2+860t+7$

I hope this helps.
Good luck.

4. I'm sorry, I"m really bad at this math stuff. So for the equations you gave me for B) and C), do I just have to plug in a number for t to solve it?

5. Originally Posted by suzdisp
I'm sorry, I"m really bad at this math stuff. So for the equations you gave me for B) and C), do I just have to plug in a number for t to solve it?
Solve them for t.

Complete the square or use the quadratic formula.

6. Originally Posted by suzdisp
I'm sorry, I"m really bad at this math stuff. So for the equations you gave me for B) and C), do I just have to plug in a number for t to solve it?
The equations given are quadratic equations. Have you been taught anything at all that's in this link?

7. I have had an overview of the quadratic equation, but I am just not getting it. It is hard because I am doing my school online, so there is really no tutor that I can sit down with and show me step by step what to do. I am one of those really dense people that needs to see EVERY step on how to solve the problems. I took my first test and failed it so now I'm trying to figure out how to do this WELL so I can pass my retake

if you have a quadratic equation in standard form
$\displaystyle ax^2+bx+c=0$ Note: it must be equal to zero

then

$\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Using this as an example if we have

$\displaystyle 6x^2+7x-5=0$

The first thing we need to do is idendify a, b and c.

a is the coeffient (the number in front of ) the $\displaystyle x^2$ term

so a = 6

b is the coeffient on the $\displaystyle x$ term so b = 7

c is the constant term -5 (note the sign comes with the number)

so evaluating the quadratic formula we get..

$\displaystyle x=\frac{-(7) \pm \sqrt{(7)^2-4(6)(-5)}}{2(6)}$

Simplifying

$\displaystyle x=\frac{-7 \pm \sqrt{49+120}}{12}=\frac{-7 \pm \sqrt{169}}{12}$

$\displaystyle x=\frac{-7 \pm 13}{12}$

so $\displaystyle x= \frac{-7+13}{12}=\frac{1}{2}$ or

$\displaystyle x=\frac{-7-13}{12}=\frac{-20}{12}=-\frac{5}{3}$

Try this on the above problems.

9. Originally Posted by suzdisp
I have had an overview of the quadratic equation, but I am just not getting it. It is hard because I am doing my school online, so there is really no tutor that I can sit down with and show me step by step what to do. I am one of those really dense people that needs to see EVERY step on how to solve the problems. I took my first test and failed it so now I'm trying to figure out how to do this WELL so I can pass my retake
Please don't take this the wrong way (I'm saying that a lot this morning!) but perhaps you should hire a personal tutor then? Just trying to think of what's best for your education.

-Dan

10. i cannot afford a personal tutor. why else would I be on here? duh just kidding!! anyways, okay. i tried to plug in my numbers for the problem and this is what I have so far:

x=-860 +/- SR 860^2 - 4(16)(7)
----------------------
2(16)
x= -860 +/- SR 739600-448
--------------
2(16)

x= -860 +/- SR 739152
----------
2(16)

x= -860 +/-860/32 ------->so 0
--
32

or -1720
------ = -53.76 Am I moving in the right direction at all to solving this problem??
32

11. anybody...is that last reply right at all???

12. Originally Posted by topsquark
$\displaystyle h = 7 + 860t - 16t^2$
-Dan
Originally Posted by suzdisp
x=-860 +/- SR 860^2 - 4(16)(7)
----------------------
2(16)
x= -860 +/- SR 739600-448
--------------
2(16)

x= -860 +/- SR 739152
----------
2(16)

x= -860 +/-860/32 ------->so 0
--
32

or -1720
------ = -53.76
32

$\displaystyle -16t^2 + 860t + 7 = 0$

So a = -16, b = 860, and c = 7.

Thus
$\displaystyle t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle t = \frac{-860 \pm \sqrt{860^2 - 4(-16)(7)}}{2(-16)}$

$\displaystyle t = \frac{-860 \pm \sqrt{739600 + 448}}{-32}$

$\displaystyle t = \frac{-860 \pm \sqrt{740048}}{-32}$

$\displaystyle t = \frac{-860 \pm 860.2604257}{-32}$

$\displaystyle t = \frac{0.2604257}{-32}$ or $\displaystyle t = \frac{-1720.2604257}{-32}$

$\displaystyle t \approx -0.008138$ or $\displaystyle t \approx 53.7581$

-Dan