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Two balls of equal masses are thrown upwards along the same vertical direction at an interval of 2 seconds with same initial velocity of 39.2m/s .Then find the height at which they collide.

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by mathukkutty

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by mathukkutty

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Displacement of 1st mass = s1 = 39.2 t - 1/2 * 9.8 * t² = 4.9 (8 t - t²)

Displacement of 2nd mass = s2 = 39.2 (t-2) - 1/2 * 9.8 * (t-2)²

= 4.9 [ 8(t-2) - (t-2)² ] = 4.9 [ 8 t - 16 - t² + 4 t - 4 ]

= 4.9 [ 8 t - t² +4 t - 20 ]

s1 = s2 => 4 t - 20 =0 => t = 5 sec

s1 = height at which they collide = 4.9 ( 8*5 - 5²) = 4.9 * 15 = 73.5 meters

Displacement of 2nd mass = s2 = 39.2 (t-2) - 1/2 * 9.8 * (t-2)²

= 4.9 [ 8(t-2) - (t-2)² ] = 4.9 [ 8 t - 16 - t² + 4 t - 4 ]

= 4.9 [ 8 t - t² +4 t - 20 ]

s1 = s2 => 4 t - 20 =0 => t = 5 sec

s1 = height at which they collide = 4.9 ( 8*5 - 5²) = 4.9 * 15 = 73.5 meters