A ground to ground projectile is at point A at t = T/3, is at point B at t = 5T/6 and reaches the ground at t = T. the difference in heights between point A and B is
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starock20kamalrock:
yup
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Let heights difference between A and B = H
Given :-
t = T/3
t = 5T/6
As we know that, Time of Flight,
T = 2u SinØ/g
u SinØ = gT/2 ------(1)
For A,
h = usinØ t - 1/2 gt²
h = usinØ × T/3 - gT²/18
h = gT²/6 - gT²/18
h = 2gT²/18
For B,
h' = usinØ t - 1/gt²
h' = gT/2 × 5T/6 - 1/2 g 25T²/36
h' = 5gT²/12 - 25gT²/72
h' = 5gT²/72
Again,
H = h - h' = 2gT²/18 - 5gT²/72
H = (8gT² - 5gT²)/72
H = 3gT²/72
H = gT²/24 Answer.......
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