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Question

# From the top of a tower 100m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with velocity of 25ms−1. Find when and where the two balls will meet. Take g=10ms−2

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Solution

## Step 1: Analysing the problem [Refer Figure]Let the two ball meet at point C at time tIf the distance covered by ball 2 is x, then the distance covered by ball 1 will be 100−x, Since the total hieght is 100m.Step 2: Equations of motion for constant accelerationSince acceleration is constant, Therefore we can apply equations of motion for both balls (considering Positive Downwards).For ball 1:u1=0 a1=g=10m/s2 s1=100−x S1=u1t+12a1t2 ⇒ 100−x=12(10)t2 ....(1)For ball 2:u1=−25m/s a2=g=10m/s2 s2=−x S2=u2t+12a2t2 ⇒ −x=−25t+1210t2 . ⇒ x=−25t−5t2 ....(2)Step 3: Solving equationsFrom equation (1) and (2) 100−25t+5t2=5t2 ⇒ t=4sPutting the value of t in equation (1) x=100−5×(4)2 =20mHence they will meet 20m above the ground after t=4s

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