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Grade: 11

                        

an elevator without a ceiling is ascending with a constant speed of 10m/s. a boy on the elevator shoots a ball directly upward, from a height of 2.0 m above the elevator floor. At this time the elevator floor is 28 m above the ground. the initial speed of the ball with respect to the elevator is 20 m/s (take g=9.8m/s²) a). what maximum height above the ground does the ball reach? b). How long does it take to return to the elevator floor

2 years ago

Answers : (1)

Aditya Gupta
2057 Points
							
this is a question of ncert.
  1. the initial speed of ball is 20+10=30m/s. so, u=30, v=0, a= –9.8. 2*-9.8*h= – 900. so, max height is 30+45.9=75.9m or approx 76m.
  2. taking the reference frame as the elevator, we have u=20, a= –9.8, v=0. so, 0=20-9.8t or t=2.04. so total time taken to return to the elevator floor is 4.08 seconds
2 years ago
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