Un proyectil se dispara con una velocidad inicial de 53 m/s. determinar el angulo de proyeccion necesario para que la altura maxima del proyectil sea igual a su alcance horizontal
yexs
![Reemplazemos~(1)~en~(3) \\ \\ t.(106cos\theta)=143,32sen\²\theta~~---\ \textgreater \ como~sabemos~[t=5,41sen\theta] \\ \\ 5,41sen\theta.(106cos\theta)=143,32sen\²\theta~~~---\ \textgreater \ resolviendo \\ \\ 573,46.cos\theta=143,32sen\theta \\ \\ \frac{573,46}{143,32}= \frac{sen\theta}{cos\theta} \\ \\ 4,0013=tg\theta \\ \\ \theta=arctg(4,0013) \\ \\](https://tex.z-dn.net/?f=+Reemplazemos%7E%281%29%7Een%7E%283%29+%5C%5C++%5C%5C+t.%28106cos%5Ctheta%29%3D143%2C32sen%5C%C2%B2%5Ctheta%7E%7E---%5C+%5Ctextgreater+%5C+como%7Esabemos%7E%5Bt%3D5%2C41sen%5Ctheta%5D+%5C%5C++%5C%5C+5%2C41sen%5Ctheta.%28106cos%5Ctheta%29%3D143%2C32sen%5C%C2%B2%5Ctheta%7E%7E%7E---%5C+%5Ctextgreater+%5C+resolviendo+%5C%5C++%5C%5C+573%2C46.cos%5Ctheta%3D143%2C32sen%5Ctheta+%5C%5C++%5C%5C++%5Cfrac%7B573%2C46%7D%7B143%2C32%7D%3D+%5Cfrac%7Bsen%5Ctheta%7D%7Bcos%5Ctheta%7D+++%5C%5C++%5C%5C+4%2C0013%3Dtg%5Ctheta+%5C%5C++%5C%5C+%5Ctheta%3Darctg%284%2C0013%29+%5C%5C++%5C%5C+ "Reemplazemos~(1)~en~(3) \\ \\ t.(106cos\theta)=143,32sen\²\theta~~---\ \textgreater \ como~sabemos~[t=5,41sen\theta] \\ \\ 5,41sen\theta.(106cos\theta)=143,32sen\²\theta~~~---\ \textgreater \ resolviendo \\ \\ 573,46.cos\theta=143,32sen\theta \\ \\ \frac{573,46}{143,32}= \frac{sen\theta}{cos\theta} \\ \\ 4,0013=tg\theta \\ \\ \theta=arctg(4,0013) \\ \\")
1 votes
Thanks 1
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.