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- Thread starter avb203796
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- #2

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I am not even really sur ehow to get started with this problem!

- #3

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y = xtantheta - [gx^2/(2v^2cos^2theta)] where x is horizontal distance covered andy is vertical distance. Use this and report if you still have problem. Also try deriving the above formula.

you must also be knowing that t = 2vsintheta/g

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- #5

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then substitute t in the equation y = vsintheta*t - 1/2g*t^2

- #6

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It may seem like a 3D motion qn but it is actually juz 2D..

For the 1st part, u noe that the ball touches the water 5s after the golfer hits the ball.. u oso noe that the height (oso known as the displacement) above the water where the golfer hits is 4m.. u oso noe that the vertical acceleration for bodies in 2D motion is only gravity ( 9.81m/s^2) and what u wan now is the initial velocity.. note: <---this is vertical.. u still nid to get use pythagoras theorm on both the horizontal and vertical velocities to get the initial velocity..

I hope this gets u start off.. the rest shld be quite easy..

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