Finding vertical velocity at highest point in trajectory

[honey.s]

Hey ya'll ... I just need clarification on this solution. Is it correct in me saying the vertical component is negative as the ball is falling to the ground? Also, I was unsure whether to use the vertical velocity or release velocity in finding the highest point of trajectory

Q: A polo ball is hit from the ground at an angle of 33 degrees upwards from the horizontal. If it has a release velocity of 30 m/s and lands on the ground, calculate the vertical and horizontal components of its release velocity and provide these answers in the answer boxes provided.

(i) What are the vertical and horizontal components?

sin33= x/30

vertical component = 16.34 (should this be negative because the ball is falling down??)

cos33= y/30

Horizontal component = 25.16

(ii)What will be the vertical velocity of the ball be at the highest point in its trajectory.

s=Final velocity - initial velocity / 2a

s= 0-(16.34)^2 / 2(9.8)

=13.6 (is it correct in me using the vertical velocity as opposed to release velocity?)

Thankyou in advance

 

[voko]

If the ball at its highest point has a positive velocity, then what prevents it from being a little higher than the "highest" a little bit later?

 

[honey.s]

If I use the release velocity, then s= 0- (30)^2 / 2 (9.8), and therefore the answer is 45.9 and at its highest. That was my initial thought, but the next question asks for the horizontal velocity of the ball at its highest point, and that's why I thought we had to use vertical and horizontal velocity respectively..

 

[voko]

You got an even higher value for vertical velocity at the apex. My argument in #2 applies.

You should think about what condition defines the highest point of the trajectory. Where is the projectile going just before the highest point, and where is it going after?