Question

A plane is flying horizontally at a height of $5000$ m with a speed of $500$ m/s. At time $t=0$ a bombshell is fired from a point on the ground exactly below the plane, at angle $\theta$ from the horizontal. The bombshell hits the plane in $10$ seconds. What is the value of $\frac{11}{2}cot\theta$ ?

Answer 1

$horizontalvelocityofthebombshellis500m/s,andtheverticaldistanceithastocoveris5000m.thus,5000=v_{y}t-\frac{g{t}^2}{2}=v_y\cdot 10-5\cdot 100v_y=550m/stan\theta =\frac{550}{500}=\frac{11}{10}\Rightarrow cot\theta =\frac{10}{11}\Rightarrow \frac{11}{2}cot\theta =5$