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A catapult hurls a stone of mass 32.0 g with velocity of 50.0 m/s at a 30.0 degree angle of elevation. a) What is the maximum height reached by the stone? b) What is its range ? c) How long has the stone been in the air when it returns to its original height? |
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Answer» y =ut− 21 gt 2 h=S y =20sin30 o ×5− 21 ×10×(5) 2 h=50−125=−75m (MINUS sign just indicates that the displacement is in DOWNWARD direction)h max = 2gu 2 sin 2 θ = 2×10(20) 2 sin 2 30 0 =5mHence, MAXIMUM height ATTAINED by projectile =h+h max =75+5=80 mplease MARK me as brainliest |
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