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Hrishant Goswami Grade: 10
         A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find (a) the maximum height reached and (b) the range of the ball. Take g = 10 m/s2. 
3 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
										Sol. u = 40 m/s, a = g= 9.8 m/s2, θ = 60° Angle of projection.

a) Maximum height h = (u^2 sin^2⁡θ)/2g = (〖40〗^2 (sin⁡〖60°〗 )^2)/(2 x 10) = 60 m
b) Horizontal range X = (u2 sin 2θ) / g = (402 sin 2(60°)) / 10 = 80 √3 m.
3 years ago
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