A plane takes off runaway that is horizonntally 915ft from a building 121ft high. what is the minimum angle of elevation of its take off to assure of going near the building if its flies in a straight line?
+1316 You have to take off at
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{121}}}{{\mathtt{915}}}}\right)} = {\mathtt{7.533\: \!110\: \!014\: \!093^{\circ}}}$$
it would be higher than this because the plane has to reach take off speed first so it will be much closer when it take off. You want to be close to it not crash into it.
+1316 You have to take off at
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{121}}}{{\mathtt{915}}}}\right)} = {\mathtt{7.533\: \!110\: \!014\: \!093^{\circ}}}$$
it would be higher than this because the plane has to reach take off speed first so it will be much closer when it take off. You want to be close to it not crash into it.
