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?

1 Answer
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(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)