Question

A projectile is shot from the ground at an angle of `pi/6` and a speed of `1 m/s`. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

Answer 1

H=0.012m and `d_h=0.044m`

Explanation:

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 dosplacement h will be zero. So applying the equation of motion under gravity we can write

`h=usinalphaxxT+1/2gT^2`
`=>0=usinalphaxxT-1/2xxgxxT^2`
where `g= "acceleration" "dueto"" gravity”`
`:.T=(2usinalpha)/g`
The horizontal displacement during this T sec is `R=ucosalpha xxT`
`R=(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`
By the problem
`u=1m/s;g=9.8m/s^2 and alpha= pi/6`
`d_h=1/2xx(1^2sin(2*pi/6)/9.8)=0.044m`
If H is the maximum height then
`0^2=u^2sin^2(alpha)-2*g*H`
`:. H= (u^2sin^2(alpha))/(2*g)=(1^2sin^2(pi/6))/(2*9.8)=0.012m`

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((0.044)^2+(0.012^2))=0.045m`