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5 Find the maximum height and range for vo = 10 m/s, and ? = 90°.

him sharma , 12 Years ago
Grade 12
anser 2 Answers
ruchi yadav
Maximum height in projectile motion ish={v_{0}^{2}\sin ^{2}(\theta ) \over {2g}}

h = (10)2sin2(900)/ 2(10)

h = 5 m


Range in projectile motion is d = \frac{v^2 \sin 2 \theta}{g}


d = (10)2 sin(1800)/10

Range =0
Last Activity: 12 Years ago
ruchi yadav
Maximum height in projectile motion ish={v_{0}^{2}\sin ^{2}(\theta ) \over {2g}}

h = (10)2sin2(900)/ 2(10)

h = 5 m


Range in projectile motion is d = \frac{v^2 \sin 2 \theta}{g}  


d = (10)2sin(1800)/10

Range =0
Last Activity: 12 Years ago
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