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


(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

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  • $\begingroup$ You're absolutely right. $\endgroup$ – philip_0008 May 29 '16 at 6:01
  • 1
    $\begingroup$ What's an "MCQ"? Always explain non-ubiquitious abbreviations the first time you use them! $\endgroup$ – ACuriousMind May 29 '16 at 9:55
  • $\begingroup$ "MCQ" is a multiple choice question. $\endgroup$ – 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.

  • $\begingroup$ So it means the correct answer is 10m/s $\endgroup$ – Mahin May 29 '16 at 6:26
  • $\begingroup$ Is there any doubt left? I suppose you can find the numeric result on your own. $\endgroup$ – ja72 May 29 '16 at 6:37

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