. The table is ft high.
A ball rolls off a table with a speed ofSuppose the ball rebounds from the floor at the same angle with which it hits the floor, but loses % of its speed due to energy absorbed by the ball on impact. Where does the ball strike the floor on the second bounce?
Enter the answer as the displacement of the ball in feet to the right of the table's edge.
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Let time of first & second flight are t 1 & t 2
2 1 g t 1 2 = h ⇒ t 1 = 3 2 . 1 7 4 2 × 3 . 5 = 0 . 4 6 6 4 4 s e c R 1 = u t 1 = 2 × 0 . 4 6 6 4 4 = 0 . 9 3 2 8 8 f t
Let speed just before rebound is v.
The path of ball will be same if ball will be thrown towards table from that point with speed v so,
R 1 = 2 1 g v 2 s i n 2 θ ( m u l t i p l i e d b y 2 1 a s o n l y h a l f f l i g h t )
As it loses 20% speed,
R 2 = g ( 5 4 v ) 2 s i n 2 θ
R 2 = 2 5 1 6 g v 2 s i n 2 θ = 2 5 3 2 R 1 ⇒ A n s = R 1 + R 2 = ( 1 + 2 5 3 2 ) R 1 = 2 5 5 7 × 0 . 9 3 2 8 8 = 2 . 1 2 7 f t