Grade 12Mechanics
maximum height reached by projectile is 4m. The horizontal range is 12m. velocity of projection?
1 Answer
Arun
Dear Student
H = (R tan Ф ) / 4
4 = (12 tan Ф ) / 4
tan Ф = 4/3 sin 2Ф = 2 tan Ф / (1 + tan² Ф) = 0.96
R = u² sin 2Ф / g
u² = 12 * g / 0.96 = 25 g /2
u = 5 √(g/2)
4 = (12 tan Ф ) / 4
tan Ф = 4/3 sin 2Ф = 2 tan Ф / (1 + tan² Ф) = 0.96
R = u² sin 2Ф / g
u² = 12 * g / 0.96 = 25 g /2
u = 5 √(g/2)
Regards
Arun (askIITians forum expert)
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