An aeroplane is flying horizontally with a velocity of 10 m/s and at a height of 1960 m. When it is vertically above a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is (ignoring air resistance)
Correct answer: B. 200 m
- A. 100 m
- B. 200 m
- C. 400 m
- D. 2 km
Explanation
200 m: Understanding Projectile Motion:The horizontal velocity of the bomb remains constant (10 m/s).We need to find the time it takes for the bomb to fall 1960 m.Then we use the time to calculate the horizontal distance traveled.Calculations:Using the equation h = (1/2)gt², where h is the height, g is acceleration due to gravity (9.8 m/s²), and t is time:1960 = (1/2) * 9.8 * t²t² = 1960 / 4.9 = 400t = √400 = 20 secondsHorizontal distance (AB) = horizontal velocity * timeAB = 10 m/s * 20 s = 200 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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