If a projectile is launched with 3m/s velocity at 60-degree angle then at the highest point its horizontal velocity is:
Correct answer: C. 1.5 m/s
- A. 3 m/s
- B. 2m/s
- C. 1.5 m/s
- D. 1.8 m/s
Explanation
At the highest point of the projectile's trajectory, its horizontal velocity will be equal to the initial horizontal velocity. Therefore, the horizontal velocity of the projectile at the highest point will be:v*cosθ = 3*cos(60) = 1.5 m/swhere v is the initial velocity and theta is the launch angle.
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Displacement, velocity and acceleration describe motion, including the horizontal and vertical components of projectile motion. The chapter connects these quantities with Newton's laws, momentum and impulse, then applies conservation of momentum to collisions, distinguishing elastic collisions from inelastic ones.
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