From the top of the tower of height400 m , a ball is dropped by a man. Simultaneously from the base of the tower, another ball is thrown up with velocity50 m/s g=10 ms-2. At what distance will they meet from the base of the tower?
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a
100 m
b
320 m
c
80 m
d
240 m
answer is C.
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Detailed Solution
For rising ball distance moved up is s=50t−12g(t2) …(1) For failing ball distance fallen and y=400−s=12g(t2) …(2) Adding, 50t=400 or t=8 sec ∴ s=50×8−12×10×64 =400−320=80 m
From the top of the tower of height400 m , a ball is dropped by a man. Simultaneously from the base of the tower, another ball is thrown up with velocity50 m/s g=10 ms-2. At what distance will they meet from the base of the tower?