Confusion about projectile [closed]

Yesterday i got a MCQ in exam that a body in projectile motion with initial velocity 20m/s creating an angle of 60 degree with horizontal. What is its velocity at max. Height

(A) 10m/s

Or

(B) 0m/s?

I think that it should be 10m/s but all friends are saying that it is 0m/s. Please tell me where am i getting it wrong. Because i think only vertical velocity is 0m/s at max. Height so the velocity of projectile would be only horizontal component of initial velocity. $V = V_i \cos(\theta)$.

closed as off-topic by ACuriousMind♦, AccidentalFourierTransform, Qmechanic♦May 29 '16 at 11:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, AccidentalFourierTransform, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

• You're absolutely right. – philip_0008 May 29 '16 at 6:01
• What's an "MCQ"? Always explain non-ubiquitious abbreviations the first time you use them! – ACuriousMind May 29 '16 at 9:55
• "MCQ" is a multiple choice question. – Mahin May 29 '16 at 22:54

You are correct. The horizontal component of velocity is constant at $v_x = V_i \cos \theta$. The vertical component varies by $v_y = V_i \sin \theta - g t$.
Together the speed at any point is $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{V_i^2 + (g t)^2 - 2 V_i (g t) \sin \theta}$$
At the top of $v_y=0$ (by definition) then $v=V_i \cos \theta$ as you mentioned.