a object is thrown upward to a certain height with a velocity of 3 M per second if the time required to reach the height is 5 seconds calculate the height at which is was thrown (hint g=10ms-2 use second equation of motion
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v^2=u^2+2gS
(3^2) = 0+ 2*10*S
9 = 0 + 20S
THERFORE S=0.45m
Hope this helps
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muskan6347:
i am not understand please explain
use newtons second law of motion which says v^2=u^2+2gS ..... v is final velocity u is initial velocity g is acceleration due to gravity and S is dispacment .... substitute the values and ull get the value of S
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*second equation of motion
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