I'm having trouble with solving this problem:
An aircraft flying at an altitude of 35,000 feet above the Atlantic Ocean with a speed of 481 knots releases a 3,200-kilogram payload. Neglecting friction, temperature and wind,
find:
a) the tangential velocities of the payload at heights :15,000 feet, 5,000 feet, and 0 feet,
b) the tangential and normal components of acceleration acting on the payload at the moment of impact with the water, and
c) its range from release and angle of impact to the horizontal (1 knot=6076ft/hr).
Any help would be appreciated.
Best Answer
Let height at which the aircraft is flying be $h$ and acceleration due to gravity be $g$. Since we are neglecting viscous forces, the following equations apply, where $v=\text{final velocity}$,$\; u=\text{initial velocity}$,$\; t=\text{time taken from release to touchdown}$,$\;s=\text{displacement},\; a=\text{acceleration},\; g=\text{acceleration due to gravity}$.( We will denote horizontal quantities with the subscript $\; \rightarrow$ and vertical ones with $\; \downarrow \;$) $$\begin{align} v&=u+at..........(1) \\ v^2 &=u^2+2as..........(2) \\ \end{align} $$ Consider vertical components first: $u_\downarrow =0;\;a_\downarrow=g;\;s=h$ $$\begin{align} \text{from (2)}: v_\downarrow &=\sqrt{u_\downarrow^2+2a_\downarrow s}\\ &=\sqrt{u_\downarrow^2+2gs}\\ &=\sqrt{0+2gh}=\sqrt{2gh}\\ \end{align} $$ $$\begin{align}\text{from (1)}:\; t=\frac{v_\downarrow-u_\downarrow}{a_\downarrow}&=\frac{v_\downarrow-u_\downarrow}{g}=\frac{\sqrt{2gh}}{g}=\sqrt{\frac{2h}{g}} \end{align}$$ Note that since $a_\rightarrow=0,\;v_\rightarrow=u_\rightarrow,\; \text{irrespective of height}$
a)$$\begin{align}& \text{At 15,000 ft, 5000 ft and 0 ft, displacement is 25000ft, 30000 ft and 35000 ft, respectively} \\ & Hence\; v_\downarrow=\sqrt{\frac{2s}{g}}=\sqrt{\frac{50000}{g}},\;\sqrt{\frac{60000}{g}},\;\sqrt{\frac{70000}{g}} \;\text{respectively.}\\ & \text{tangential velocity}=\sqrt{v_\downarrow^2+v_\rightarrow^2}=\sqrt{v_\downarrow^2+u_\rightarrow^2}\\ &\;\text{at an angle, below the horizontal, of}\;\arctan \frac{v_\downarrow}{u_\rightarrow} \end{align}$$
b)$\;\text{tangential component}\;=\;a_\rightarrow=0,\;\text{and}\;\text{normal component}=g\;$
c)$\;\text{range}=u_\rightarrow \cdot t$