A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point with no air resistance. What was the angle of launch?



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Posted on (Answer #1)

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

`rArr t=(usinalpha)/g`

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)






`=52.5` degrees

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