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If a particle is projected with an angle 45° with velocity 50m/s. What will be the max.range of the projectile.

Aditya Pathak , 7 Years ago
Grade 12th pass
anser 4 Answers
Mohd Mujtaba

Last Activity: 7 Years ago

Answer is 250m. We know maximum range =u^2sin2theta /g. Put values we get maximum range=250m. Hope you understand. Thank☺☺☺☺

DIWAKAR PANDEY

Last Activity: 7 Years ago

The max range will be u^2×2sin x cos x /g by putting values in this formula the range will come 250 m please approve my ans if you have understood the concept of using the formula. Thank u

ashwin

Last Activity: 6 Years ago

max.range=[usin2(Q)]/g,(where ‘Q’ is the angle of projection)
                 =[502xsin(2x45)]/9.8
                 =[2500x1]/9.8         …....(sin90=1)
                 =255.102 m
 

Sai Disha

Last Activity: 6 Years ago

Angle - 45
Velocity - 50 
We have to substitute the values of velocity and angle ( theta ) 
u - 50
Theta - 45

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