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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.

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

Grade:12th pass

4 Answers

Mohd Mujtaba
131 Points
6 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
168 Points
6 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
29 Points
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
17 Points
5 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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