Question

A projectile is shot from the ground at an angle of `pi/6` and a speed of `1 /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 `=0.0005678m`

Explanation:

Resolving in the vertical direction `uarr^+`

The initial velocity is `u_0=(1/9)sin(1/6pi)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-((1/9)sin(1/6pi))^2/(-2g)`

`h_y=(1/9sin(1/6pi))^2/(2g)=0.0001575m`

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-(1/9)sin(1/6pi))/(-9.8)=0.0057s`

Resolving in the horizontal direction `rarr^+`

The velocity is constant and `u_x=(1/9)cos(1/6pi)`

The distance travelled in the horizontal direction is

`s_x=u_x*t=(1/9)cos(1/6pi)*0.0057=0.0005455m`

The distance from the starting point is

`d=sqrt(h_y^2+s_x^2)=sqrt(0.0001575^2+0.0005455^2)=0.0005678m`