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.
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.
For first ball
s = ut +
h = 180 - 6u -(1)
for second boy
h = 1.5u + 11.25 -(2)
from (1) & (2)
180 - 6u = 1.5u + 11.25
7.5u = 180 - 11.25
h = 180 - 6u
= 180 - 6 * 22.5
45m
for third ball
t2 = 9
t = 3 sec
Similar Questions for you
Please find the solution below:
[h] = ML2T-1
[E] = ML2T-2
[V] = ML2T-2C-1
[P] = MLT-1
According to question, we can write
10 =
Average speed
(d) Initial velocity
Final velocity
Change in velocity
Momentum gain is along
Force experienced is along
Force experienced is in North-East direction.
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physics ncert solutions class 11th 2023
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