Question

A projectile is shot from the ground at an angle of `pi/4` and a speed of `1/2 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

`0.0152 m`

Explanation:

If initial velocity of the projectile is `u`, and projected at an angle `theta` w.r.t horizontal,its time`(t)` to reach the highest point is `(u sin theta)/g` (using `v= u-at`)

Maximum height reached = `(u^2 sin theta)/(2g)`

Given, `u= 0.5 m/s` and `theta = 45`

So,maximum height`(h)` reached by it is `0.0088 m`

In that time (t) horizontally it will reach a distance `(r)` of `(u cos 45)*(u sin theta)/g` (see value of t given previously)
Or,`0.0125 m`

So,distance of projectile will be `sqrt` `(r^2+h^2)` i.e `0.0152 m`(as the distance values are so small hence we can think the arch formed by the projectile in its pathway to be a straight line)