Question

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?

Answer 1

40m/s

Step-by-step 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