Question

A projectile is shot from the ground at an angle of `pi/8` and a speed of `3/5 m/s`. When the projectile is at its maximum height, what will its distance from the starting point be?

Answer 1

The distance is `=0.013m`

Explanation:

Resolving in the vertical direction `uarr^+`

The initial velocity is `u_0=3/5sin(1/8pi)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-(3/5sin(1/8pi))^2)/(-2g)`

`h_y=(3/5sin(1/8pi))^2/(2g)=0.0028m`

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)=(0-3/5sin(1/8pi))/(-9.8)=0.023s`

Resolving in the horizontal direction `rarr^+`

The velocity is constant and `u_x=3/5cos(1/8pi)`

The distance travelled in the horizontal direction is

`s_x=u_x*t=3/5cos(1/8pi)*0.023=0.013m`

The distance from the starting point is

`d=sqrt(h_y^2+s_x^2)=sqrt(0.0028^2+0.013^2)=0.013m`