A particle projected with a speed of 50 m/s has a range that is thrice its greatest height. Find the range
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A particle projected with a speed of 50 m/s has a range that is thrice its greatest height. Find the range
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Answer:
R = 240 m
Explanation:
Given,
R = 3H ; u = 50m/s
W.K.T, R = u^2sin2ø/g
H = u^2 sin^2ø/2g
R = 3H
u^2(2sinøcosø)/g = 3u^2sin^2ø/2g
2cosø = 3sinø/2
4/3 = tanø
tanø = 4/3 = OP/AD, HYP = root 4^2 + 3^2 = 5
sinø = 4/5
cosø = 3/5
R = u^2(2sinøcosø)/g
R = 50(50)(2)(4/5)(3/5)/10
R = 100(2)(4)(3)/10
R = 240 m