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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.

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2 Answers 2

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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)$\;\text{range}=u_\rightarrow \cdot t$

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This answer is not going to go into detail but outline the steps needed. If you need additional explanation, please ask in the comments:

Procedure for (a):

  1. The tangential velocity is just the velocicy represented as a vector. (Velocity is always tangent to the path).
  2. You know that the horizontal component of velocity is constant due to zero friction assumption.
  3. The vertical component of velocity will have to be computed from $v=at$. You can determine the time from the $d=\frac{1}{2}at^2$ formula. Make sure you pay attention to the sign of the velocity.

Procedure for (b):

  1. Use the velocity at 0 feet from (a) to define the direction of the tangent.
  2. Determine the horizontal acceleration. Hint: the horizontal velocity is constant.
  3. Determine the vertical acceleration. Hint: What is the acceleration due to gravity?
  4. Use the dot product to project the acceleration onto the tangential direction. Hint: $v\cdot a = \|v\|\|a\|\cos(\theta)$. Second hint: What is $\|a\|\cos(\theta)$?
  5. Compute the normal component by removing the tangent component.

Procedure for (c):

  1. Knowing $t$ at impact and knowing the horizontal speed easily gives the answer.
  2. The angle of impact can be deduced from the velocity vector components. Hint use the $\tan^{-1}$ function.
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