A body is thrown from the surface of the Earth at an angle `alpha` to the horizontal with the initial velocity `v_0`. Assuming the air drag to be negligible, find:
(a) the time of motion,
(b) the maximum height of ascent and the horizontal range, at what value of the angle `alpha` they will be equal to each other,
(c) the equation of trajectory `y(x)`, where y and x are displacements of the body along the vertical and the horizontal respectively,
(d) the curvature radii of trajectory at its initial point and at its peak.

Answer

(a) `tau=2(v_0//g)sin alpha`;
(b) `h=(v_0^2//2g)sin^2alpha`, `l=(v_0^2//g)sin2alpha`, `alpha=76^@`;
(c) `y=xtanalpha-(g//2v_0^2cos^2alpha)x^2`;
(d) `R_1=v_0^2//gcosalpha`, `R=(v_0^2//g)cos^2alpha`.

Explanation

The body thrown in air with velocity `v_0` at an angle `alpha` from the horizontal lands at point P on the Earth's surface at same horizontal level (figure). The point of projection is taken as origin, so, `Deltax=x` and `Deltay=y`
(a) From the Eq. `Deltay=v_0._vt+1/2w_yt^2`
`0=v_0sin alphatau-1/2g tau^2`
As `tau !=0`, so, time of motion `tau=(2v_0sin alpha)/(g)`
(b) At the maximum height of ascent, `v_y=0`
so, from the Eq. `v_y^2=v_(0y)^2+2w_yDeltay`
`0=(v_0sin alpha)^2-2gH`
Hence maximum height `H=(v_0^2sin^2alpha)/(2g)`
During the time of motion the net horizontal displacement or horizontal range, will be obtained by the equation
`Deltax=v_(0x)t+1/2w_xtau^2`
or, `R=v_0cosalphatau-1/2(0)tau^2=v_0cos alpha tau=(v_0^2sin 2alpha)/(g)`
when `R=H`
`(v_0^2sin^2alpha)/(g)=(v_0^2sin^2alpha)/(2g)` or `tan alpha=4`, so, `alpha=tan^-14`
(c) For the body, `x(t)` and `y(t)` are
`x=v_0cos alphat` (1)
and `y=v_0sin alpha-1/2g t^2`
Hence putting the value of t from (1) into (2) we get,
`y=v_0sin alpha ((x)/(v_0cos alpha))-1/2g((x)/(v_0cos alpha))^2=xtanalpha-(gx^2)/(2v_0^2cos^2alpha^')`
Which is the sought equation of trajectory i.e. `y(x)`
(d) As the body thrown in air follows a curve, it has some normal acceleration at all the moments of time during it's motion in air.
At the initial point `(x=0, y=0)`, from the equation:
`w_n=v^2/R`, (where R is the radius of curvature)
`g cos alpha=(v_0^2)/(R_0)` (see figure) or `R_0=(v^2)/(g cos alpha)`
At the peak point `v_y=0, v=v_x=v_0 cos alpha` and the angential acceleration is zero.
Now from the Eq. `w_n=v^2/R`
`g=(v^2cos^2alpha)/(R)`, or `R=(v_0^2cos^2alpha)/(g)`
We may use the formula of curvature radius of a trajectory `y(x)`, to solve part (d),
`R=([1+(dy//dx)^2]^(3//2))/(|d^2y//dx^2|)`