A ball is dropped from a height. If it takes 0.2s to cross the last 6m before hitting the ground, find the height from which it is dropped. (answer should be 48.5m) you try
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Answered by
73
Height = H
Total time = T
H = 0.5gT^2
H - 6 = 0.5g(T - 0.2)^2
0.5gT^2 - 6 = 0.5g(T^2 + 0.04 - 0.4T)
0.5gT^2 - 6 = 0.5gT^2 + 0.02g - 0.2gT
-6 = 0.02g - 0.2gT
T = (0.02g + 6) / (0.2g)
T = [(0.02 * 10) + 6) / (0.2 * 10)
T = 3.1 s
H = 0.5gT^2
= 0.5 * 10 * 3.1^2
= 48.05 m
Total time = T
H = 0.5gT^2
H - 6 = 0.5g(T - 0.2)^2
0.5gT^2 - 6 = 0.5g(T^2 + 0.04 - 0.4T)
0.5gT^2 - 6 = 0.5gT^2 + 0.02g - 0.2gT
-6 = 0.02g - 0.2gT
T = (0.02g + 6) / (0.2g)
T = [(0.02 * 10) + 6) / (0.2 * 10)
T = 3.1 s
H = 0.5gT^2
= 0.5 * 10 * 3.1^2
= 48.05 m
ehbdhjxhknh:
thanka for your instruction . I have got it
bro, T=0.02g+6/(0.2g) ilts wrong
it will be, -6=2(-0.2t+0.02)
i took common from it
then , -6/2+0.2t=0.02
t=0.02/0.2+3
then t=3.1
in this way we get it
You got 3.1 and I also got 3.1 how you're correct and I'm wrong? Of course both of us did correct.
oooh! sorry , you did by taking the formula of "g"
Answered by
21
this question is there in my material
thanks
thanks
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