Question

A projectile is shot from the ground at an angle of `pi/4` and a speed of `3 /8 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 `0.007m`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=3/8*sin(1/4pi)`

Acceleration is `a=-g`

At the maximum height, `v=0`

We apply the equation of motion

`v=u+at`

to calculate the time to reach the greatest height

`0=3/8sin(1/4pi)-g*t`

`t=3/8*1/g*sin(1/4pi)`

`=0.027s`

Resolving in the horizontal direction `rarr^+`

We apply the equation of motion

`s=u_x*t`

`=3/8cos(1/4pi)*0.027`

`=0.007m`