Question

# A ball is dropped from a height of 20 meters and at the same time another ball is thrown up from the ground with the speed of 20 m/s. When and where will the balls meet?

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Solution

## Let us assume that the two balls meet at a height 'h' after time 't' above the ground. For the ball dropped from the top of the tower: Distance covered by the ball is (20 - h) m Here u = 0 ; s = (20 - h) m and g = 9.8 ms-2 Using the second equation of motion, → s = ut + 1/2 at2 or, 20 - h = 0 × t + 1/2 at2 or, 20 - h = 4.9t2 ............(Equation 1) For the ball thrown vertically upwards: u = 20 ms-1 ; s = h and g = -9.8ms-2 (-ve value of g since thrown upwards) s = ut + 1/2 at2 h = 20 × t + 1/2(-9.8)t2 or, h = 20t - 4.9t2 ............(Equation 2) Adding equations (1) and (2), we get 20 - h + h = 4.9t2 + 20t - 4.9t2 On solving we get t = 1 s Substituting t = 1 second in equation 1, we get 20 - h = 4.9 × (1)2 or, h = 20 - 4.9 = 15.1 m Thus, the two balls meet at a height 15.1 m from the ground after 1 s.

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