a body is dropped from height H. the time taken to cover half of the journey is
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If it takes time “T” to reach the ground then,
H = 0.5gT²
T = √(2H/g)
If it takes time “t” to travel first half of the journey then,
0.5H = 0.5gt²
H = gt²
t = √(H/g)
Time taken to cover second half of the journey is = T - t
= √(2H/g) - √(H/g)
= √(H/g)×(√(2) - 1)
H = 0.5gT²
T = √(2H/g)
If it takes time “t” to travel first half of the journey then,
0.5H = 0.5gt²
H = gt²
t = √(H/g)
Time taken to cover second half of the journey is = T - t
= √(2H/g) - √(H/g)
= √(H/g)×(√(2) - 1)
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1
Hi
As it is dropped , u =0
Total height is h but half distance means h/2
Using
S=ut+1/2gt sq
h/2=1/2×10t
H/10=t
Hope it is helpful
As it is dropped , u =0
Total height is h but half distance means h/2
Using
S=ut+1/2gt sq
h/2=1/2×10t
H/10=t
Hope it is helpful
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