A 2 kg object is projected at 20 m/s at 53° above the horizontal. What is the maximum height reached? (g = 9.8 m/s², neglect air resistance)
Correct answer: C. 13.1 m
- A. 4.8 m
- B. 6.4 m
- C. 13.1 m
- D. 10.2 m
Explanation
To find the maximum height, we first resolve the initial velocity into its vertical component using vy = v sin 53°. Since sin 53° ~ 0.8, the vertical velocity becomes vy= 20 x 0.8 = 16 m/s. The maximum height of a projectile is given by H = vy2/ 2g . Substituting the values, we get H = 162/ 2 x 9.8 = 256/19.6 ~ 13.1 m. Thus, the object rises to a maximum height of about 13.1 m.
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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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