2020-03-30T11:41:34-04:00
A water sprinkler bursts from the ground with a velocity of 8.46m/s at an angle of 45°. Find:
I) The magnitude height
II) The range of the sprinkler
III) Total time the water took to reach ground
1
2020-04-05T14:33:06-0400
i)
h = v 2 sin 2 45 2 g = 8.4 6 2 sin 2 45 2 ⋅ 9.8 = 1.83 m h=\frac{v^2\sin^2{45}}{2g}=\frac{8.46^2\sin^2{45}}{2\cdot9.8}=1.83\ m h = 2 g v 2 sin 2 45 = 2 ⋅ 9.8 8.4 6 2 sin 2 45 = 1.83 m
ii)
R = v 2 sin ( 2 ⋅ 45 ) g = 8.4 6 2 9.8 = 7.30 m R=\frac{v^2\sin{(2\cdot45)}}{g}=\frac{8.46^2}{9.8}=7.30\ m R = g v 2 sin ( 2 ⋅ 45 ) = 9.8 8.4 6 2 = 7.30 m
iii)
t = 2 v sin 45 g = 2 8.46 sin 45 9.8 = 1.22 s t=2\frac{v\sin{45}}{g}=2\frac{8.46\sin{45}}{9.8}=1.22\ s t = 2 g v sin 45 = 2 9.8 8.46 sin 45 = 1.22 s
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