a ball is thrown up with a velocity 29.23ms^-1 distance travelled in the last second of upward motion is
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we know the ball has an initial velocity, u = 29ms^-1
and that there is a constant decelleration of 9.8ms^-2
at the maximum height, velocity, v = 0
so we can use the equation V^2 = u^2 + 2as
which implies (29)^2 + 2x(-9.8)s = 0
so 29^2 = 2x9.8xs
so s = 29^2 / (2x9.8)
so s = approx 42.9 m
hope this helps
If this isn't clear enough tell me and I'll try to make it a bit neater and easier to understand
and that there is a constant decelleration of 9.8ms^-2
at the maximum height, velocity, v = 0
so we can use the equation V^2 = u^2 + 2as
which implies (29)^2 + 2x(-9.8)s = 0
so 29^2 = 2x9.8xs
so s = 29^2 / (2x9.8)
so s = approx 42.9 m
hope this helps
If this isn't clear enough tell me and I'll try to make it a bit neater and easier to understand
NithishReddy6332:
1) 2.3m 2) 6m 3) 9.8m and 4) 4.9m
it is 4.9 answer
how
could you plz explain me the answer
it was the approx value see in answer
u kept 42.9 and it is too far from 4.9
can i ask you some more questions related to physics
u can
sure
i kept questions with a photo
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