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?

### 1 Answer | Add Yours

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` ``

`=(u^2sin^2alpha)/(2g)`

Horizontal distance covered in this time: `(ucosalpha*usinalpha)/g`

By symmetry of the parabolic motion, total horizontal range=`(2*ucosalpha*usinalpha)/g`

=u^2sin2alpha/g

By the condition of the problem, `(u^2sin2alpha)/(g)=111` --- (i)

and `(u^2sin^2alpha)/(2g) =72.3` --- (ii)

---------------

(ii)/(i)

---------------

`tanalpha=2*72.3/111`

`alpha=arctan(2*72.3/111)`

`=52.5` degrees

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes