Question

A cannon ball is launched at an 18 degree angle with a speed of 25 m/s. What is it’s horizontal velocity at the peak of its arc?

Answer 1

Given that the cannon ball is launched at an 18 degree angle

Initial speed = 25 m/s

So the initial velocity in the horizontal direction (say v_x) will be

$v_x=25cos18\degree$

The horizontal component of velocity will be constant throughout the motion since there is no acceleration in the x-direction, only acceleration due to gravity in y-direction.

So at the peak of its arc the velocity will be the same to its initial value, i.e.

$v_{peak,x}=25cos18\degree=23.78\text{ m/s}$

Result: Horizontal velocity at peak of its arc will be 23.78 m/s.