1.

M=-1,point (-2,-3apon 2)​

Answer»

Step-by-step explanation:

maximum height can be attained only when the PROJECTILE is LAUNCHED at 90°

threrfore ,

H= u^2/2g

25×2×9.8= u^2

490=u^2

7√10=u

then for range ,

the maximum range OCCURS at 45°

R = u^2/g

R = 490/9.8

R = 50 m



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