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b) A ball is thrown up and caught back by the thrower after 8 second. Calculate (i) the velocity with which the ball was thrown up, (ii) the maximum height attained by the ball and (iii) distance covered by the ball after 2 second of the start of motion. (Take g = 10 m/s2 |
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Answer» ken to go to the top = 8/2 s = 4 s Acceleration = G = - 10 m/s² (Upward)(i) FINAL velocity = 0 m/s, g = 10 m/s².We know, or ⇒ ⇒ (ii) We know, or, ⇒ ⇒ (iii) We know, or, ⇒ ⇒ .∴ Initial velocity of the ball was of 40 m/s, maximum height ATTAINED by it is 80 m and distance travelled by the ball after 2 seconds of motion is 60 m. |
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