Please solve this...?
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Velocity at highest point = 1/2 mv^2cos^2theta -----------(i)
Given mv^2 = 1/2 -------------(ii)
On solving (i) and (ii) we get
mv^2cos^2theta = mv^2/2
cos theta = 45 degrees...
Given mv^2 = 1/2 -------------(ii)
On solving (i) and (ii) we get
mv^2cos^2theta = mv^2/2
cos theta = 45 degrees...
siddhartharao77:
Correct me if am wrong with the answer.
no your answer is right
Answered by
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It's quite simple question
but it looks like a toughest one.
For this , we have,
initial velocity= u
initial kinetic energy =1/2mv^2
now at the highest point only horizontal component of velocity will be present hence
final velocity ,v= ucostheta
final kinetic energy = 1/2m(ucostheta)^2
According to the question,
1/2× 1/2mu^2= 1/2m(ucostheta)^2
u^2/2 = (ucostheta)^2
costheta= 1/√2 , hence theta is 45 °
but it looks like a toughest one.
For this , we have,
initial velocity= u
initial kinetic energy =1/2mv^2
now at the highest point only horizontal component of velocity will be present hence
final velocity ,v= ucostheta
final kinetic energy = 1/2m(ucostheta)^2
According to the question,
1/2× 1/2mu^2= 1/2m(ucostheta)^2
u^2/2 = (ucostheta)^2
costheta= 1/√2 , hence theta is 45 °
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