A ball is projected vertically upwards with an initial velocity of u goes to maximum height H before coming to ground what is value of h
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We know that a certain height is reached two times in a vertical projection.(One while going upward,one while going back downward)
let the height reached in time t be h
maximum height=H
difference in height=H-h
now H is reached in T and h is reached in t time.So H-h is reached in (T-t) time ...(1)
while going back down H-h is reached in (T-t)seconds
So,
H-h=u(T-t) + ½g(T-t)2
but u =0 (velocity at max height is zero)
So,
H-h = ½g(T-t)2
h = H – ½g(T-t)2 or H – ½g(t-T)2
let the height reached in time t be h
maximum height=H
difference in height=H-h
now H is reached in T and h is reached in t time.So H-h is reached in (T-t) time ...(1)
while going back down H-h is reached in (T-t)seconds
So,
H-h=u(T-t) + ½g(T-t)2
but u =0 (velocity at max height is zero)
So,
H-h = ½g(T-t)2
h = H – ½g(T-t)2 or H – ½g(t-T)2
Answered by
13
Answer:
We know that a certain height is reached two times in a vertical projection.(One while going upward,one while going back downward)
let the height reached in time t be h
maximum height=H
difference in height=H-h
now H is reached in T and h is reached in t time.So H-h is reached in (T-t) time ...(1)
while going back down H-h is reached in (T-t)seconds
So,
H-h=u(T-t) + ½g(T-t)2
but u =0 (velocity at max height is zero)
So,
H-h = ½g(T-t)2
h = H – ½g(T-t)2 or H – ½g(t-T)2
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