with initial vel. 700 m/s, what is the angle for a launch to reach 12,500 m? (horizontally)

Question

aaronq · Answer

Just take into account the component of the vector in the x direction.

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$v_{i~x}=cos\theta v_i$

lucaz · Answer

yeah, but dont we need the time?

lucaz · Answer

time taken to travel the given distance?

aaronq · Answer

hm yeah, you're right. 

Maybe it's possible to use the vertical velocity, solve for time, and plug that into the x-motion equation?

lucaz · Answer

I tried to find t when v(y) is zero, but then what is the angle... I dont know

aaronq · Answer

$\sf v_f=v_i+a\Delta t$

$\sf 0=sin\theta v_i-g\Delta t\rightarrow \Delta t=\dfrac{sin\theta v_i}{g}$

Then:

$\sf\Delta x=cos\theta v_i\Delta t +\dfrac{1}{2}(0)(\Delta t)^2$

$\sf\Delta x=cos\theta v_i(\dfrac{sin\theta v_i}{g})=\dfrac{cos\theta sin \theta v_i}{g} $

but then you'd have to use some trig

aaronq · Answer

that's what i was thinking, but i'm not entirely sure that assuming that the final velocity is 0 in the y direction is sound.

aaronq · Answer

I forgot to square the velocity btw

lucaz · Answer

my book has this formula with sin and cos.. I will try again, thank you

aaronq · Answer

no problem, good luck!