From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in __________ s.

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6 months ago

For first ball

 s = ut +   1 2 a t 2

h = u * 6 1 2 * 1 0 * 6 * 6

h = 6 u 1 8 0      

h = 180 - 6u                     -(1)

for second boy

h = u t 1 2 a t 2

h = u * 1 . 5 + 1 2 * 1 0 * 1 . 5 * 1 . 5             

h = 1.5u + 11.25               -(2)

from (1) & (2)

180 - 6u = 1.5u + 11.25

7.5u = 180 - 11.25

u = 1 6 8 . 7 5 7 . 5 = 2 2 . 5    

h = 180 - 6u

= 180 - 6 * 22.5

45m

for third ball

h = u t + 1 2 a t 2

h = 0 * t + 1 2 * 1 0 t 2     

4 5 = 5 t 2          

t2 = 9

t = 3 sec

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physics ncert solutions class 11th 2023

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