# Tricky projectile motion question, Thanks (1 Viewer)

#### Drdusk

##### π
Moderator
$\bg_white \Delta x_1 = 100t$

$\bg_white \Delta x_2 = 100\cos \bigg( \dfrac{\pi}{6}\bigg) t$

$\bg_white \text{Subtracting these two equations gives you how far apart the balls land which is 600m.}$

$\bg_white \text{Find out the time of flight 't' from the above and then just use it with the equations for vertical motion to find H}$

#### shmitler

##### New Member
$\bg_white \Delta x_1 = 100t$

$\bg_white \Delta x_2 = 100\cos \bigg( \dfrac{\pi}{6}\bigg) t$

$\bg_white \text{Subtracting these two equations gives you how far apart the balls land which is 600m.}$

$\bg_white \text{Find out the time of flight 't' from the above and then just use it with the equations for vertical motion to find H}$
hmmm, i tried that and still got answer wrong, these were my steps
100t-86.6t=600, therefore t=44.78
then i plugged t=44.78 into y=1/2gt^2+vsinθt +H (with v=100 and θ =30)
since we have the time of flight the value of y will be zero leading to the only unknown being H.
Thanks

#### Drdusk

##### π
Moderator
hmmm, i tried that and still got answer wrong, these were my steps
100t-86.6t=600, therefore t=44.78
then i plugged t=44.78 into y=1/2gt^2+vsinθt +H (with v=100 and θ =30)
since we have the time of flight the value of y will be zero leading to the only unknown being H.
Thanks

#### Nash__

##### New Member
This seems too hard. Wouldn’t the time of flight be different for each projectile??

#### shmitler

##### New Member
hahahaha really different, answer is 88 m, im getting around -7000

#### fan96

##### 617 pages
For both balls we have

$\bg_white \begin{cases} y_1(t) &= -\frac{9.8}{2}t^2+100\sin(\pi/6)t+h \\ x_1(t) &=100\cos(\pi/6)t\end{cases}$

$\bg_white \begin{cases} y_2(t) &= -\frac{9.8}{2}t^2+h \\ x_2(t) &=100t\end{cases}$

Solve $\bg_white y_1(t) = 0$ and $\bg_white y_2(t) = 0$ to get the time of flight for each ball (call these $\bg_white t_1$ and $\bg_white t_2$).

Obviously the ball shot straight horizontally will hit the ground sooner (and will travel less), so you need to solve

$\bg_white x_1(t_1)-x_2(t_2) = 600$ (not the other way around!)

for $\bg_white h$.

I calculated $\bg_white h \approx 81.2$.

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