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A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point with no...
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Let the angle of launch, with respect to the horizontal be alpha and the initial firing velocity, u
Constant horizontal velocity: ucosalpha
Initial vertical velocity: usinalpha
Time to reach the peak: t
From the laws of motion: 0=usinalpha-gt
Vertical distance covered in this time: `h=usinalphat-(1/2g*(usinalpha)/g)^2` ``
Horizontal distance covered in this time: `(ucosalpha*usinalpha)/g`
By symmetry of the parabolic motion, total horizontal range=`(2*ucosalpha*usinalpha)/g`
By the condition of the problem, `(u^2sin2alpha)/(g)=111` --- (i)
and `(u^2sin^2alpha)/(2g) =72.3` --- (ii)
Posted by llltkl on September 27, 2013 at 4:50 AM (Answer #1)
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