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For the missile aimed to achieve the maximum range of 8500 km, what is the maximum altitude reached in the trajectory? Answer in units of km.

2006-11-07 14:39:59 · 3 answers · asked by Anonymous in Science & Mathematics Physics

3 answers

Maximum range is reached if the missile is launched at an angle of 45 degrees from the horizontal. This can be proved using calculus. Thus the problem can be solved using this known fact and the given range of 8500km.

The vertical component of the velocity =vsin45. Since the angle is equal to 45 degrees, the horizontal component is the same, i.e. vsin45.

The time it takes the missile to travel 8500km can be computed using the basic formula: time=distance/velocity. Thus

t =8500/vsin45 Eq. 1). Where vsin45 is the horizontal component of the initial velocity of the missile.

Now use the formula: v^2-u^2=2as, and substitute known values, as follows:

(vsin45)^2-0=2*9.8s Equation 2

We take vsin45 as the final vertical component of the velocity; u=0, being the initial velocity at the top of the flight of the missile; a the acceleration of gravity equal to 9.8m/s^2, and s is the maximum altitude reached in the trajectory.

(You also have to know that the initial vertical component of the velocity upward is equal to the final vertical velocity downward of the free falling body dropped from the top of the flight of the missile.)

We now use another formula for s:

s=ut+1/2at^2
s=0+1/2*9.8t^2 Equation 3)

But from equaton 1) t=8500/vsin45 This is for the entire flight of the missile. The time for a free falling body to reach the ground from the top of the light of the missile is only half this. Thus divide the given expression by 2 and substitute this in equation 3):

s=1/2*9.8*[1/2*8500/v sin45]^2

substitute the value of (vsin45)^2 given in equation 2):

s=1/2*9.8*1/4*8500^2/2*9.8s
s=1/16*72250000/s
s^2=4,515,625
s=2,125 km

2006-11-08 21:53:07 · answer #1 · answered by tul b 3 · 0 0

Okay, the maximum range are key words. You are assuming no air resistance so the maximum range occurs when you launch the missile at 45 degrees.

You know that there is no acceleration in the x direction and the acceleration in the y direction is -9.8 m/s^2

Hope that helps.

2006-11-07 22:57:07 · answer #2 · answered by ic3d2 4 · 0 0

you can solve this question using vector components.
Dx= 8500km, Dy=?, ay (g)= 9.8m/s. Do you have the angle that the missile was launched at? or the initial velocity of the missile? You need more than just this one piece to solve this question.

2006-11-07 22:45:53 · answer #3 · answered by Anonymous · 0 0

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