Answer:
C) 19 m/s
Explanation:
The motion of the cannonball is a projectile motion, which consists of 2 independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion (constant acceleration) along the vertical direction
As a result, we have the following:
- The horizontal velocity of the cannonball remains constant during the motion, and it is given by
[tex]v_x = u cos \theta[/tex]
where
u = 25 m/s is the initial velocity
[tex]\theta=40^{\circ}[/tex] is the angle
Substituting,
[tex]v_x = (25)(cos 40^{\circ})=19 m/s[/tex]
- The vertical velocity keeps changing during the motion due to the acceleration of gravity. However, at the top of the trajectory, the vertical velocity is zero:
[tex]v_y = 0[/tex]
This means that at the top of its path, the cannonball has only horizontal velocity, so its velocity is
C) 19 m/s
