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joja [24]
3 years ago
9

Back to 'Energy & Work'

Physics
1 answer:
mixer [17]3 years ago
3 0

Answer:

Explanation: Two possible solutions

in the absence of friction

mgh = ½mv²

h = v²/2g = 24.4²(2(9.81) = 30.344...  = 30.3 m

if enough rolling friction exists to keep the ball from sliding, but ignoring air resistance and assuming the ball is already rolling smoothly at the start

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½((2/3)mR²)(v/R)²

2gh = v² + (2/3)v²

2gh = (5/3)v²

h = (5/6)v²/g

h = (5/6)24.4²/9.81 = 50.5742 = 50.6 m

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xf = 5.68 × 10³ m  

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solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0                            yi = 0

xf = ?                            yf = ?

vxi =  vicosθ               vyi = visinθ  

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xf = xi + vxi × t + 0.5 × ax × t²     ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m  

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ²     ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s)  ²

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3 years ago
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