# projectile

• Jan 2nd 2017, 03:20 AM
markosheehan
projectile
a particle is projected from a point on the horizontal ground with a speed u inclined to a angle a to the horizontal ground. the particle is at a height of h above the horizontal ground at times t1 and t2, show that t1*t2=2h/g
this is what ive tried h=usinat-4.9t^2 i tired putting that into the -b forumula but got no where
• Jan 2nd 2017, 04:36 AM
skeeter
Re: projectile
For a projectile launched and landing at ground level, the total time of motion is $t_f=\dfrac{2v_0\sin(a)}{g} = \dfrac{2v_{0y}}{g}$

Since the projectile is at the same height, $h$, at times $t_1$ and $t_2$, then $t_1-t_0=t_f-t_2 \implies t_1+t_2=t_f+t_0$.

Assuming launch occurs at $t_0=0$ ...

$\color{red}{t_1+t_2=\dfrac{2v_{0y}}{g}}$

$(t_1+t_2)^2 = t_1^2+2t_1t_2+t_2^2= \dfrac{4v_{0y}^2}{g^2} \implies$

$\color{red}{t_1^2+t_2^2=\dfrac{4v_{0y}^2}{g^2}-2t_1t_2}$

Now add the two equations for height ...

$h=v_{0y} \cdot t_1 - \dfrac{1}{2}gt_1^2$
$h=v_{0y} \cdot t_2 - \dfrac{1}{2}gt_2^2$
-------------------------------------------------------
$2h = v_{0y}(t_1+t_2)-\dfrac{1}{2}g(t_1^2+t_2^2)$

Substitute the expressions for $\color{red}{t_1+t_2}$ and $\color{red}{t_1^2+t_2^2}$ and solve for $t_1t_2$.
• Jan 3rd 2017, 11:30 AM
markosheehan
Re: projectile
when i am subbing it in do you know what t1t2 stands for . i cant find an expression for it
• Jan 3rd 2017, 11:40 AM
markosheehan
Re: projectile
i am trying to sub it in this is where i get to

2h=usina(2usina/g) -4.9(4u^2 sina^2-2t1t2) dont know what t1t2 is. do you know if im doing it right
• Jan 3rd 2017, 12:04 PM
skeeter
Re: projectile
Quote:

Originally Posted by markosheehan
when i am subbing it in do you know what t1t2 stands for . i cant find an expression for it

$t_1 \cdot t_2$ is just the product of the two times the projectile is at height $h$ ...

I worked out two expressions for you ...

$\color{red}{t_1+t_2=\dfrac{2v_{0y}}{g}}$ and $\color{red}{t_1^2+t_2^2 = \dfrac{4v_{0y}^2}{g^2} - 2}\color{blue}{t_1t_2}$

I also summed the two equations for same height at the two times ...

$2h = v_{0y}\color{red}{(t_1+t_2)} - \dfrac{1}{2}g\color{red}{(t_1^2+t_2^2)}$

substitute ...

$2h = v_{0y} \cdot \color{red}{\dfrac{2v_{0y}}{g}} - \dfrac{1}{2}g \cdot \color{red}{\left(\dfrac{4v_{0y}^2}{g^2} - 2\color{blue}{t_1t_2}\right)}$

$2h = \dfrac{2v_{0y}^2}{g} - \dfrac{2v_{0y}^2}{g} + g(t_1 \cdot t_2)$

$2h = \cancel{\dfrac{2v_{0y}^2}{g}} - \cancel{\dfrac{2v_{0y}^2}{g}} + g(t_1 \cdot t_2)$

$\dfrac{2h}{g} = t_1 \cdot t_2$

... which is what the problem asked you to show.
• Jan 6th 2017, 01:27 AM
markosheehan
Re: projectile
just in case you want to see it i came up with a much easier way to work this out let t be the 2 times t1 and t2 the 2 equations are h=ut-.5gt^2 and h=ut-.5gt^2 add this together. 2h=2ut-gt^2 rearrange gt^2-2ut+2h=0 divide by g t^2-2ut/g+2h/g t1 and t2 are the roots so t1*t2=-2h/g
• Jan 6th 2017, 03:01 AM
HallsofIvy
Re: projectile
?? You can't "let t be the two times t1 and t2". They are two different times, not the same time.
• Jan 6th 2017, 03:06 AM
markosheehan
Re: projectile
T can be 2 different times as it's a quadratic. I just should not of named it that way. It's still a valid way of working out the question