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Two balls are thrown from the top of a cliff of unknown height with equal initial speed v. = 10m/s. One is projected at an angle 0 = 45° above horizontal and second is projected at the same angle below horizontal. What is the difference between maximum horizontal distances travelled by the projectiles (in m)?

User Pesche
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Final answer:

The maximum horizontal distances travelled by the projectiles thrown at complementary angles of 45 degrees with respect to the horizontal will be equal. The vertical motion does not affect the horizontal range, leading to no difference in the distances travelled.

Step-by-step explanation:

To solve this physics problem, we use the concepts of projectile motion. Since the two balls are thrown with the same initial speed but in opposite directions with respect to the horizontal, the vertical component of their velocity will be the same, and they both will reach the same maximum height. This is because the vertical motion is independent of the horizontal motion and is only affected by gravity, which acts the same on both balls.

The horizontal distances (ranges) will differ due to their differing horizontal velocity components, but since they are thrown with the same speed and complementary angles with respect to the horizontal (45 degrees above and 45 degrees below), their horizontal ranges will be equal. This is because the horizontal range of a projectile is a function of the initial speed, the angle of projection and the height from which it is thrown - the initial vertical position doesn't affect the horizontal range if we ignore air resistance and assume the same endpoint height as the start. In this case, we are throwing from a cliff to the same endpoint height (ground level), assuming that both projectiles land at ground level. One ball is thrown upwards and the other downwards, but they both will have the same initial horizontal velocity because cos(45°) = cos(-45°). Hence, the difference between their maximum horizontal distances will be zero.

User RollRoll
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