Question

A projectile is shot from the ground at an angle of `pi/4` and a speed of `5 /8 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 dosplacement h will be zero. So applying the equation of motion under gravity we can write
`h=usinalpha xxT+1/2gT^2`
`=>0=uxxT-1/2xxgxxT^2`
,where `g="accelerationdueto""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 question
`u =5/8m=5/8*100=62.5cm and alpha=pi/4`

`d_h=1/2xx((62.5)^2sin(2*pi/4)/980)=2cm`

If H is the maximum height then
`0^2=u^2sin^2(alpha)-2*g*H`
`:. H= (u^2sin^2(alpha))/(2*g)`
`=((62.5)^2sin^2(pi/4))/(2*980)=1cm`

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(2^2+1^2)=sqrt5=2.24cm`