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For the ballistic missile aimed to achieve the maximum range of 7000 km, what is the...
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High School Teacher
If the missile is projected at a speed of u, with an angle x above the horizontal, then the horizontal and vertical velocities are:
ucosx and vsinx and the horizontal distance and height attained by the missile in time t are utcosx, and utsinx -(1/2)gt^2 respectively, where g is the acceleration due to gravitation.
The missile looses its ininitial velocity every second due to gravitational pull attains its maximum height at time when its vertical velocity is zero. So usinx-gt =0 when t = usinx/g.
So the height attained at this time = ut-(1/2)gt^2= u(usinx/g)sinx-(1/2)g(usinx/g)^2= (u^2*sin^2 x)/(2g)
When horizontal distance = 7000km, the missile is on the ground or its height from the ground is utsinx-(1/2)gt^2 = 0. Or
t(2usinx-gt) = 0. So t= 0 or t=2usinx/g. The horizontal distance covered at this time of 2usinx/g seconds is u(2usinx/g)cosx.= 2u^2sin^2 x/g. The horizontal distance u^2*sin(2x)/g is maximum when 2x=90 or x=45 degree.
So when x=45 degree, the maximum horizontal distance is u^2/g = 7000km. Or u^2 = 14000g
Therefore the maximum height attained by the missile for the maximum horizontal distance = u^2sin^2x/g = (14000g)*(sin45)^2/(2g) = 14000g(1/2)/(2g) = 3500km.
Posted by neela on February 17, 2010 at 1:09 PM (Answer #1)
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