Question

A projectile is shot from the ground at an angle of `( pi)/3` and a speed of `6 m/s`. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

Let the velocity of projection of the object be u with angle of projection `alpha` with the horizontal direction.
The vertical component of the velocity of projection is `usinalpha` and the horizontal component is `ucosalpha`

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical displacement h will be zero. So applying the equation of motion under gravity we can write
`h=usinalpha xxT+1/2gT^2` `=>0=uxT-1/2xxgxxT^2`
where `g="acceleration due to gravity"`

`:.T=(2usinalpha)/g`
The horizontal displacement during this T sec is `R=ucosalpha xxT=(u^2sin(2alpha))/g`

The time t to reach at the peak is half of time of flight (T)
So `t=1/2*T=(usinalpha)/g`
The horizontal displacement during time t is

`d_h=1/2xx(u^2sin(2alpha))/g`

If H is the maximum height then

`0^2=u^2sin^2(alpha)-2*g*H`

`:. H= (u^2sin^2(alpha))/(2*g)`

So the distance of the object from the point of projection when it is on the peak is given by

`D = sqrt(d_h^2+H^2)`

`=sqrt((u^2sin^2(2alpha))/(2g)^2+(u^4sin^4alpha)/(2g)^2`

`=u/(2g)sqrt((sin^2(2alpha))+(u^2sin^4alpha)`

`=6/(2*9.8)sqrt(sin^2((2pi)/3)+6^2sin^4(pi/3)`

`=3/(9.8)sqrt(3/4+(36*9)/16)`

`=3/(9.8)sqrt(3/4+81/4)`

`=3/(9.8)sqrt21m~~1.4m`