a ball projected vertically upwards from the top of a tower reaches the ground in 'a' seconds.if it is projected veritcally downwards from the top of same tower with same velocity it reaches the ground in "b" seconds.if it falls freely from the top of same tower,what is the time taken to reach the ground in terms of 'a' and 'b'.
Asked by architsrivastava02 | 3rd Nov, 2010, 10:07: AM
h = -ua + (ga2/2) ..... (1) [using newton's second equation]
= = ub + (gb2/2)...... (2)
Gives: u = g(a-b)/2
putting back in (1), h = g(a2 + b - a) / 2
Now, Without any initial velocity, let the time taken to reach the ground be t.
then, h = gt2/2
t = √(2h/g) = √(a2 + b - a)
Hope that satisfies your query.
Answered by | 4th Nov, 2010, 10:57: AM
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