A ball is projected at 45°; its horizontal range is 10m. The velocity of projection is
Correct answer: C. 10 m/s
- A. 5 m/s
- B. 15 m/s
- C. 10 m/s
- D. 20 m/s
Explanation
The horizontal range of a projectile is given by R = (v^2 * sin(2θ)) / g, where v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. Given R = 10m and θ = 45°, solving for v gives v = 10 m/s.Incorrect Options Explanation:(a) 5 m/s: This is half of the correct answer and doesn't account for the full velocity required for the given range.(b) 15 m/s: This value is higher than the correct answer and would result in a greater range.(d) 20 m/s: This value is not consistent with the given range and angle of projection.
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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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