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Grade: 12
        
in fround to ground projectile motion, the maximum height is 4m, the horizontal range is 12m, what is the velocity of projection ?
one year ago

Answers : (1)

Arun
22780 Points
							Dear Pradeep
maximjum hight is 4 m rang is 12m
height=u²sin²θ/2g=4
u²sin²θ=80
range=u²sin2θ/g
u²sin2θ=120
u²sin²θ=80
120/2 sinθcosθ*sin²=80
60/ sinθcosθ=80
tan θ=4/3
sinθ=4/5
u²sin2θ=120
u²*16/25=120
u=5 root120/4
u=13.6 m/s
one year ago
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