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?

Jan 17, 2018

$0.0152 m$

Explanation:

If initial velocity of the projectile is $u$, and projected at an angle $\theta$ w.r.t horizontal,its time$\left(t\right)$ to reach the highest point is $\frac{u \sin \theta}{g}$ (using $v = u - a t$)

Maximum height reached = $\frac{{u}^{2} \sin \theta}{2 g}$

Given, $u = 0.5 \frac{m}{s}$ and $\theta = 45$

So,maximum height$\left(h\right)$ reached by it is $0.0088 m$

In that time (t) horizontally it will reach a distance $\left(r\right)$ of $\left(u \cos 45\right) \cdot \frac{u \sin \theta}{g}$ (see value of t given previously)
Or,$0.0125 m$

So,distance of projectile will be sqrt $\left({r}^{2} + {h}^{2}\right)$ 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)