Question

A projectile is shot from the ground at an angle of `(5 pi)/12` and a speed of `7 /3 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 `=0.30m`

Explanation:

Resolving in the vertical direction `uarr^+`

The initial velocity is `u_0=(7/3)sin(5/12pi)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-((7/3)sin(5/12pi))^2/(-2g)`

`h_y=(7/3sin(5/12pi))^2/(2g)=0.26m`

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-(7/3)sin(5/12pi))/(-9.8)=0.23s`

Resolving in the horizontal direction `rarr^+`

The velocity is constant and `u_x=(7/3)cos(5/12pi)`

The distance travelled in the horizontal direction is

`s_x=u_x*t=(7/3)cos(5/12pi)*0.23=0.14m`

The distance from the starting point is

`d=sqrt(h_y^2+s_x^2)=sqrt(0.26^2+0.14^2)=0.30m`