Dear Student, 5 Time of flight, T=2usinθg=2×392×sin 6009 8=2×392×32×9 8=69 3 secHorizontal range, R=u2sin 6009 8=392×392×32×9 8=13579 27 mMaximum height, H=u2sin 2θ2×9 8=392×392×34×2×9 8=5880 m Regards

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Dear Student,
5.

Time of flight, T = \frac{2 usinθ}{g} = \frac{2 \times 392 \times \sin 60^{0}}{9.8} = \frac{2 \times 392 \times \sqrt{3}}{2 \times 9.8} = 69.3 \sec Horizontal range, R = \frac{u^{2} \sin 60^{0}}{9.8} = \frac{392 \times 392 \times \sqrt{3}}{2 \times 9.8} = 13579.27 m Maximum height, H = \frac{u^{2} \sin^{2} θ}{2 \times 9.8} = \frac{392 \times 392 \times 3}{4 \times 2 \times 9.8} = 5880 m

Regards

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