Q32 of 35 Page 55

From the top of tower 100m in height, a ball is dropped, and at the same instant a ball is projected vertically upward with a velocity of 25 ms–1 Find when and where the two balls will meet. (g = 9.8 ms–2)

Given: Height of tower = 100 m

Let the two balls meet at a height h form ground.


Ball 1 dropped from the top of the tower.


Ball 2 thrown up with a velocity of 25 ms–1.



We will use these three equations:


..1


..2


..3


Where v = Final Velocity t = Time


u = Initial Velocity s = Distance


a = Acceleration


Ball 1:


s = 100 – h; a = g = 9.8 ms–2; u = 0;


Using equation 2:




..4


Ball 2


s = h; a = –g = –9.8 ms–2; u = 25 ms–1;


Using equation 2:



..5


Using 4 and 5,



t = 4 s ..6


Using 4 and 6


h = 100 – 4.9 × 16


h = 21.6 m


So the two balls meet at a distance 21.6 m above the ground.


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