If a ball is launched horizontally at 40 m/s

from a bridge, what will be the magnitude of

its horizontal velocity after 3 seconds?

40m/s

Explanation:

The horizontal component of velocity remains constant because there are no external forces in that direction

By applying motion equations, V = U + at

where,

v - final velocity u - initial velocity a-acceleration t - time

v = u + at

As no force act on the ball (we neglect air resistance here) no acceleration is seen,

So v = u = 40m/s