Question

A projectile is shot from the ground at an angle of `( pi)/3` and a speed of `9 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

The distance is `=4.74m`

Explanation:

Resolving in the vertical direction `uarr^+`

The initial velocity is `u_0=9sin(1/3pi)ms^-1`

Applying the equation of motion

`v^2=u^2+2as`

At the greatest height, `v=0ms^-1`

The acceleration due to gravity is `a=-g=-9.8ms^-2`

Therefore,

The greatest height is `h_y=s=(0-(9sin(1/3pi))^2)/(-2g)`

`h_y=(9sin(1/3pi))^2/(2g)=3.1m`

The time to reach the greatest height is `=ts`

Applying the equation of motion

`v=u+at=u- g t`

The time is `t=(v-u)/(-g)=(0-9sin(1/3pi))/(-9.8)=0.795s`

Resolving in the horizontal direction `rarr^+`

The velocity is constant and `u_x=9cos(1/3pi)`

The distance travelled in the horizontal direction is

`s_x=u_x*t=9cos(1/3pi)*0.795=3.58m`

The distance from the starting point is

`d=sqrt(h_y^2+s_x^2)=sqrt(3.1^2+3.58^2)=4.74m`

Answer 2

`~~ 4.733` m

Explanation:

The maximum height the projectile reaches is
`H = {u^2 sin^2 alpha}/{2g} = {81times 3/4}/{2 times 9.8} ~~3.1` m

At this point, the horizontal distance traveled is half of the horizontal range `D = u^2/{ g} sin alpha cos alpha ~~ 3.578` m

The distance from the starting point is, then

`sqrt{H^2+D^2} ~~ 4.733` m