A ball is thrown up with 20 ms-1 at an angle of 60° with x-axis. The horizontal velocity of the ball at the top position is:
Correct answer: C. 10 ms-1
- A. 0 ms-1
- B. 20 ms-1
- C. 10 ms-1
- D. 16 ms-1
Explanation
In projectile motion, the horizontal component of velocity remains constant throughout the motion, as there is no horizontal acceleration (ignoring air resistance). The formula for the horizontal component of velocity is Vx = Vcosθ. Given that V = 20 ms-1 and θ = 60°, we calculate Vx = 20cos60° = 20 × 0.5 = 10 ms-1. Therefore, the correct answer is 10 ms-1.Option A incorrectly assumes the horizontal velocity is zero, which applies only to the vertical component at the peak. Option B incorrectly suggests the initial velocity is the same as the horizontal component. Option D provides an arbitrary value that does not result from the given calculations.
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