Two balls are thrown simultaneously, (A vertically upwards with a speed of 20 metre per second)

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Answer

Text Solution

Solution : Suppose the balls collide at a height y from the ground after time t from the start <br> For downward motion of ball B <br> `40-y =20t +(1)/(2)" gt"^(2)" ….(i)"` <br> For upward motion of ball A <br> `y=20t -(1)/(2)" gt"^(2)" …..(ii)"` <br> Adding equations (i) and (ii), we get 40 = 40t or t = 1s Now from equation (ii), we get `y=20 xx1-(1)/(2) xx9.8xx1^(2) =15.1 m` <br> <img src="//d10lpgp6xz60nq.cloudfront.net/physics_images/VMC_NEET_XI_PHY_MOD_01_C03_SLV_043_S01.png" width="80%">