Question

A projectile is shot from the ground at an angle of `( pi)/3` and a speed of `7/5 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 `=0.115m`

Explanation:

Resolving in the vertical direction `uarr^+`

The initial velocity is `u_0=(7/5)sin(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-((7/5)sin(1/3pi))^2/(-2g)`

`h_y=(7/5sin(1/3pi))^2/(2g)=0.075m`

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-(7/5)sin(1/3pi))/(-9.8)=0.124s`

Resolving in the horizontal direction `rarr^+`

The velocity is constant and `u_x=(4/9)cos(1/6pi)`

The distance travelled in the horizontal direction is

`s_x=u_x*t=(7/5)cos(1/3pi)*0.124=0.087m`

The distance from the starting point is

`d=sqrt(h_y^2+s_x^2)=sqrt(0.075^2+0.087^2)=0.115m`