a person sees an object on a Tree at height of 40 m and at a distance of 60 M with what velocity he should throw an arrow at an angle of 45° so that it may hit the object? Take g =10m/sec square
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First assume velocity to be 'u' then use trajectory equation Y= Xtan45 - g(X)^2 /(u^2 (cos45)^2).
Where X =60
Y=40
Answer is 60 m/s
Where X =60
Y=40
Answer is 60 m/s
JunaidMirza:
You wrote trajectory equation incorrect. It should be g(X)^2 /(2u^2 (cos45)^2)
Oh sorry typing error
Why you didn't edited your answer?
Answered by
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y = xTanθ - gx² / (2u²Cos²θ)
40 = 60Tan45 - (10 × 3600) / (2u² Cos²45)
40 = 60 - 36000 / (2u²)
36000 / (2u²) = 20
u² = (36000 / 40)
u = 30 m/s
Velocity of arrow should be 30 m/s
40 = 60Tan45 - (10 × 3600) / (2u² Cos²45)
40 = 60 - 36000 / (2u²)
36000 / (2u²) = 20
u² = (36000 / 40)
u = 30 m/s
Velocity of arrow should be 30 m/s
thanks i got it
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