# Thread: Homework question about projectile:

1. ## Homework question about projectile:

A projectile is launched with speed $V_o$ at an angle $\alpha_{o}$ above the horizontal. The launch point is at a height h above the ground.

Show that if air resistance is neglected, the horizontal distance that the projectile travels before striking the ground is:

$x = \frac{V_{o}\cos{\alpha_{o}}}{g}(V_{0}\sin{\alpha_{ o}} + \sqrt{V_{o}^2\sin^2{\alpha_{o}} + 2gh})$

Verify that if the launch point is at ground level so that h = 0 this is equal to the horizontal range R = $\frac{{V_{0}}^2\sin{2\alpha_{o}}}{g}$ (derived from the example in my texbook)

Ok.....

the projectile is the first case and the second case is the angle below h ...

and im very confused at which equations to gather here:
$x = V_{o}\cos{\alpha_o}$

$y = (V_{o}\sin{\alpha_o})t - \frac{1}{2}gt^2$

$V_x = V_o\cos{\alpha_o}$

$V_y = V_o\sin{\alpha_o}$

alpha_o is initial angle
help needed for physics pros here (Especially topsquark)

2. Originally Posted by ^_^Engineer_Adam^_^
A projectile is launched with speed $V_o$ at an angle $\alpha_{o}$ above the horizontal. The launch point is at a height h above the ground.

Show that if air resistance is neglected, the horizontal distance that the projectile travels before striking the ground is:

$x = \frac{V_{o}\cos{\alpha_{o}}}{g}(V_{0}\sin{\alpha_{ o}} + \sqrt{V_{o}^2\sin^2{\alpha_{o}} + 2gh})$

Verify that if the launch point is at ground level so that h = 0 this is equal to the horizontal range R = $\frac{{V_{0}}^2\sin{2\alpha_{o}}}{g}$ (derived from the example in my texbook)

Ok.....

the projectile is the first case and the second case is the angle below h ...

and im very confused at which equations to gather here:
$x = V_{o}\cos{\alpha_o}t" alt="x = V_{o}\cos{\alpha_o}t" />

$y = (V_{o}\sin{\alpha_o})t - \frac{1}{2}gt^2$

$V_x = V_o\cos{\alpha_o}$

$V_y = V_o\sin{\alpha_o}$

alpha_o is initial angle
help needed for physics pros here (Especially topsquark)
I take it you have noticed the range formula doesn't work here.

I'm not going to take the time to fill in all the steps right now (I'll get back to you later if you need that done) but you've got (most of) your information correct and I'll give you some steps to work through.
First, answer the following question. Your coordinate system likely has the projectile at a height h above the origin at the start of the problem:
Yes! If so then your equation is $y = h + (V_{o}\sin{\alpha_o})t - \frac{1}{2}gt^2$ and your final height y is 0. Solve for t.

No! Then you have your origin at the point where the projectile was fired, so the final height of the projectile when it hits the ground, y, is -h. So $-h = (V_{o}\sin{\alpha_o})t - \frac{1}{2}gt^2$. Solve for t.

You now have the time of flight. Now plug that t value into the equation
$x = V_{o}\cos{\alpha_o}t$
which is the corrected x equation.

-Dan

3. ## Thanks TS

quadratic formula...
$t = \frac{-V_o\sin{\alpha_o} +- \sqrt{{V_o}^2\sin^2{\alpha_o}+2gh}}{-g}$

ok im stuck with this
and the plus / minus sign

brb after class:

4. The projectile is launched at an initial velocity of $v_0$ so $v_{x_0}=v_0\cos\alpha_0$ and $v_{y_0}=v_0\sin\alpha_0$. We will substitute these at the end to safe space.

We are told that the acceleration in the horizonal direction is $0$ so the velocity in the horizontal direction is:
$a_x=0$
$v_x=\int a_x dt = \int 0 dt = 0 + C \Longrightarrow v_x=v_{x_0}$ because $C=v_{x_0}$

and the position in the horizontal direction is:
$x=\int v_xdt=\int v_{x_0}dt = v_{x_0}t + C \Longrightarrow x=v_{x_0}t$ because $C=0$

The acceleration in the vertical direction is $-g$ so the velocity in the vertical direction is:
$a_y=-g$
$v_y = \int -g dt=-gt + C \Longrightarrow v_y=-gt + v_{y_0}$ because $C=v_{y_0}$

and the position in the vertical direction is:
$y=\int v_ydt=\int -gt + v_{y_0}dt = -\frac{1}{2}gt^2 + v_{y_0}t + C \Longrightarrow y = -\frac{1}{2}gt^2+v_{y_0} +h$ because $C=h$.

Because the projectile hits the ground when $y=0$, we use the quadratic equation to solve for $t$.
$t=\frac{-v_{y_0} \pm \sqrt{v_{y_0}^2-4\cdot (-\frac{1}{2})gh}}{2\cdot (-\frac{1}{2})g}=$ $\frac{v_{y_0}+\sqrt{v_{y_0}^2+2gh}}{g}$ we may ignore the minus because we need a positive time and to do that, we need a positive numerator which we can only get if the plus or minus is a plus.

We want to know the range of the projectile so we substitute the $t$ we just got.
$x = v_{x_0}t = v_{x_0}\left(\frac{v_{y_0} + \sqrt{v_{Y_0}^2+2gh}}{g}\right)$.

And finally substituting the initial values of the velocities we have:
$x=\frac{v_0\cos\alpha_0}{g}$ $\left(v_0\sin\alpha_0 + \sqrt{v_0^2 \sin^2 \alpha_0 + 2gh}\right)$

I know this is a long and for the most part unnecessary explanation but I feel that if I have the time, I should be as clear as possible. Also, it helps me practice latex. Hope it helps.

5. ## rualin

THanks a million!!!

6. I just hope it helped. That is exactly how I would have worked the problem if it had been assigned to me. Also, thanks to topsquark for giving me enough time to type out the solution. We newbies have a reputation to make.

7. Originally Posted by rualin
I know this is a long and for the most part unnecessary explanation but I feel that if I have the time, I should be as clear as possible. Also, it helps me practice latex. Hope it helps.
Not unnecessary. Engineer_Adam apparently needed the extra steps spelled out for him. So you did what was needed. And as far as giving time for the newbies, I was beat last night so I didn't log on. (Really it isn't me that's a problem in this area. You need to speak with Jhevon, who has a nack for answering questions quickly 24/7. I apparently need more sleep than he does. )

-Dan